How to solve difficult Algebra problems in a few simple steps 2

Use End State Analysis and Three variable zero sum principle to solve the problem in a few simple steps

How to solve algebra problems in a few simple steps 2

The problems involving Algebra are taught in schools and also form an important part of most of the competitive job tests such as SSC CGL, Bank POs etc.

Causes behind difficulties with Algebra

First : Though the topic of Algebra is based on a small set of concepts, dealing with abstract symbolic variables poses first level of difficulty in forming the problem definition in comparison to problems involving numbers.

Second: Furthermore, based on the small number of building blocks of concepts, Algebraic expressions can be made to appear very complex posing the second level of difficulty to the student in understanding and solving such problems.

Third: Identification of useful patterns in complex algebraic expressions is an essential skill in solving Algebra problems, while not everyone is adept in recognizing useful patterns that is not easily visible. This forms the third level of difficulty with problems on Algebra in general.

Some of these important and useful patterns may systematically be compiled into a set of rich concepts on top of the basic concepts, for solving such problems easily in future.

Fourth: Knowledge of basic and rich concepts is though not enough - one needs to have basic problem solving skills using powerful general problem solving strategies and techniques.

Fifth: Lastly, it is a fact of life and Algebra that even after intensive skill enhancing activities, you may face a totally new type of problem in the actual question paper. You need to be able to bridge the final gap, as we say, to be in full control of any domain, in this case, Algebra.

Why it is interesting and how the problems are dealt with

The aspect that we like in Algebra lies in its posing the mysteries that can be unraveled with a bit of effort by non-mathematicians and interested common folks - it is not usually out of reach.

Concept layers

In our last article on Algebra, for the first time we have presented the concepts (without proof) that are useful for solving problems in this area. Like any subject or activity area, the important and useful concepts fall in two layers - the basic concept layer that everyone knows, and the rich concept layer that is derived from the basic concepts and actual problem solving experience and added to the layer incrementally.

We will repeat this time our basic concept layer as before for your convenience and more importantly will extend the rich concept layer significantly.

The problem solving concept and process layer uses these subject concept layers to solve any problem in the domain efficiently.

Remember, to us it is not problem solving any which way - it is efficient problem solving at low cost (in this case time) .

Basic concepts

The basic operations involved in Algebra are none other than all the basic arithmetic operations, but on abstract symbolic variables and expressions, not on numbers.

The more important relationships in Algebra forming the basic concept layer of Algebra are the following.

$(a + b)^2 = a^2 + 2ab + b^2$, in the form of square of sum of two variables.

$(a - b)^2 = a^2 - 2ab + b^2$, negative counterpart of the sum of two variables.

$a^2 - b^2 = (a + b)(a - b)$, this is one of the most useful algebraic relationships.

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, in the form of cube of sum of two variables.

$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$, this cube form is used frequently and it is better to remember it as a basic concept.

$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, negative counterpart of cube of sum of two variables.

$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$, which again is useful.

Derived from the cubes of sums we get the next set of basic relationships that are frequently used.

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, a very useful relationship to be remembered and used in the right places.

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, negative counterpart and equally useful.

Similarly it pays to know the similar expressions in three variables.

$(a + b + c)^2 $

$\hspace{1mm} = a^2 + b^2 + c^2 + 2(ab + bc + ca)$,

$(a + b + c)^3 $

$\hspace{1mm} = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c +a)$,

$a^3 + b^3 + c^3 - 3abc $

$\hspace{1mm} = (a + b + c) \times{}$

$\hspace{10mm} (a^2 + b^2 + c^2 - ab - bc - ca)$.

Rich concepts

1. Three variable zero sum principle

The rich concept of three variable zero sum principle is,

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$ which is a very useful concept.


Proof of three variable zero sum principle:

$a + b + c = 0$,

$(a + b + c)^3 $

$ = (a + b)^3 + $

$\hspace{10mm} 3(a + b)c(a + b + c) + c^3$

$ = a^3 + 3ab(a + b) + b^3 + c^3$, $a + b + c = 0$ eliminates second term of the last expression.

$ = a^3 - 3abc + b^3 + c^3 = 0$, again from $a + b + c = 0$, we use, $a + b = -c$,

Or, $a^3 + b^3 + c^3 = 3abc$, a compact and highly useful result.

To identify this highly useful result we have given the concept a name of Three variable zero sum principle and included it in our rich concept layer.

This powerful rich concept has been used in this session itself when solving a problem later.

Second: Chained equation treatment technique

When we meet a chained equation of the form, $a^x = b^y = c^z$, which we call as a chained equation, a strategy of adding another artificial equality proves generally to be very useful as it frees each expression in the chained equation and allows independent pairwse comparison and useful result derivation.

The new equality is introduced in the form,

$a^x = b^y = c^z = q$, from which we can form independent equations, $a^x = q$, $b^y = q$ and $c^z = q$, all in terms of $q$.

You may refer to the effective use of this technique in the first session on how to solve difficult Algebra problems in a few simple steps.

Third: Base equalization technique

We have found application of this profound technique in widely different areas, of Maths and in real life problems as well.

You may explore its use in indices problems, in first session on how to solve difficult Algebra problems in a few simple steps, fraction problems and in diverse areas.

Fourth: Principle of inverses

This is another very useful concept set. You may refer to its detailed treatment in our article on inverses.

Briefly one of the useful results of principle of inverses in Algebra is,

If $x + \displaystyle\frac{1}{x} = n$, where $n$ usually is a suitable positive integer, we can always derive similar expressions in sum of inverses for powers 2, 3 and beyond. The basic advantage with this type of expressions of inverses results from the variable $x$ disappearing when the two inverses are multiplied together, $x\times{\displaystyle\frac{1}{x}} = 1$.

Example problem:

If $x + \displaystyle\frac{1}{x} = 2$, find $x^3 + \displaystyle\frac{1}{x^3}$.


$x + \displaystyle\frac{1}{x} = 2$,

Squaring we get,

$x^2 + \displaystyle\frac{1}{x^2} + 2 = 4$,

Or, $x^2 + \displaystyle\frac{1}{x^2} = 2$,

So, $x^3 + \displaystyle\frac{1}{x^3}$

$\hspace{5mm}= \left(x + \displaystyle\frac{1}{x}\right)\left(x^2 - 1 + \displaystyle\frac{1}{x^2}\right)$

$\hspace{5mm}= 2$.

The principle of inverses can also be applied in solving real life problems.

Fifth: Substitution technique or principle of representative

When in complex expressions we observe some of the component expressions appear unchanged throughout the problem, our first action will always be to replace these more complex component expressions by single variables.

This immediately simplifies the clutter and visual complexity of the problem, and brings into focus commonly used results that were not visible earlier due to the clutter.

We will show effective use of this technique later in this session itself while solving a problem.

This again is a general problem solving principle and is applied unknowingly in many diverse real life problem areas.

Sixth: Principle of collection of friendly terms

This is a very useful technique applied especially in Algebra with great effectiveness. It simply says,

In a complex algebraic expression with many individual terms, collect together the terms in small groups, so that each group can take up a new meaningful existence, thus simplifying the whole expression in one glorious sweep.

We have already highlighted the use of this concept in our first session on how to solve difficult Algebra problems in a few simple steps.

Additionally we will show its effective use again in this session itself while solving a problem later.

Seventh: Input transformation technique

Though this sounds simple, many tricky problems can be elegantly solved by transforming the given expression in well known symmetric forms that bear direct relations to the end state expression.

We will show effective use of this technique in this session itself while solving a problem in a short while.

Let us now get on with our elegant problem solving process.

Problem 1.

If $x = a(b - c)$, $y = b(c - a)$ and $z = c(a - b)$, then $\left(\displaystyle\frac{x}{a}\right)^3 + \left(\displaystyle\frac{y}{b}\right)^3 + \left(\displaystyle\frac{z}{c}\right)^3$ is,

  1. $3xyzabc$
  2. $\displaystyle\frac{xyz}{3abc}$
  3. $\displaystyle\frac{3xyz}{abc}$
  4. $\displaystyle\frac{xyz}{abc}$


First stage Problem analysis:

The very first thing that hits you is the large number of six variables and long expressions. Looking closely at the end state expression to be evaluated, we find a pattern of sum of three cubes of $\displaystyle\frac{x}{a}$, $\displaystyle\frac{y}{b}$ and $\displaystyle\frac{z}{c}$.

If you think of expanding these terms, you know from experience that you won't reach anywhere soon. Rather your first effort would be (as you are habituated now in End state analysis that says in a nutshell, look for similarity between the end state and the given state) to try to find presence of the compound variables $\displaystyle\frac{x}{a}$, $\displaystyle\frac{y}{b}$ and $\displaystyle\frac{z}{c}$ in the given the three expressions.

As you are searching now for a specific pattern, you find it in no time.

First stage input transformation

$x = a(b - c)$,

Or, $\displaystyle\frac{x}{a} = (b - c)$


$y = b(c - a)$,

Or, $\displaystyle\frac{y}{b} = (c - a)$, and

$z = c(a - b)$,

Or, $\displaystyle\frac{z}{c} = (a - b)$

First clue:

We found our first breakthough, the compound variables as observed in the end state expressions also appear in the given expression. So we get rid of this comlexity of form immediately by using substitution or principle of representation. We assume,

$p =\displaystyle\frac{x}{a}$,

$q = \displaystyle\frac{y}{b}$, and

$r = \displaystyle\frac{z}{c}$.

By this stratagem we rewrite our whole problem in much simpler form.

Problem simplification

If $p = (b - c)$, $q = (c - a)$ and $r = (a - b)$, then find the value of $p^3 + q^3 + r^3$. What a transformation, isn't it?

Second stage problem analysis

Dictated by End state analysis approach again we analyze the transformed end state sum of cubes expression and the simplified given expressions together.

As we look at the sum of cubes expression, we want so much to have the zero sum in three variables in the given expressions. We know from our knowledge of three variable zero sum principle that the result out of this situation would be very simple.

With this intention of finding a three variable zero sum expression in p, q and r, we examine the three given expressions and indeed we find it as,

$p + q + r = (b - c) + (c - a) + (a - b) = 0$

The final result

From our rich concept on the principle we already know that, if $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$, and so in our case,

$p^3 + q^3 + r^3 = 3pqr = \displaystyle\frac{3xyz}{abc}$.

Answer: Option c: $\displaystyle\frac{3xyz}{abc}$.

Key concepts used:

  • By the use of End State Analysis on the target end state expression, identification of the expression as a sum of cubes of compound variables.
  • Using end state analysis, searching for the presence of the compound variables in the given expressions and finding them.
  • Using input transformation technique and substitution technique transformation of the input expressions as well as the end state expression involving compound variables into a simple form involving simple variables.
  • Again using end state analysis searching for the presence of three variable zero sum expression and identifying it.
  • Getting the three variable zero sum expression by using Principle of collection of friendly terms, just by summing up the given expressions together.
  • Use of the simple result of well known three variable zero sum principle.

Thus we reached our elegant solution.

Summary of deductive steps:

Substituting for $\displaystyle\frac{x}{a} = p$, $\displaystyle\frac{y}{b} = q$, and $\displaystyle\frac{z}{c} = r$, the given problem is simplified to (both given expressions and end state expression),

If $p = (b - c)$, $q = (c - a)$ and $q = (a - b)$ then find $p^3 + q^3 + r^3$.

As $p + q + r = 0$, from three variable zero sum principle we know that,

$p^3 + q^3 + r^3 = 3pqr$,

Or, the target expression $ = 3pqr = \displaystyle\frac{3xyz}{abc}$.


When we put the deduction steps together, we get only a few simple but intelligently analyzed steps. In fact, when we face this question in a competitive test under time pressure, we do not actually deduce these steps on paper. We do it all in mind in twenty seconds at most.

This is always possible if we are aware and experienced in using our basic and rich concepts on the topic Algebra, and also the right problem solving strategies along with identification of useful patterns at all stages of the deduction process.

The detailed explanation embodies the thinking involved in solving the problem. We call this as our Deductive Reasoning which is much more than usual Reasoning.

Problem 2.

If $a^3 - b^3 - c^3 = 0$, then $a^9 - b^9 - c^9 - 3a^3b^3c^3$ is,

  1. $0$
  2. $-\displaystyle\frac{11}{4}$
  3. $-\displaystyle\frac{49}{4}$
  4. $\displaystyle\frac{49}{2}$

Problem analysis

The target expression together with the given expressions remind us of a familiar result of three variable zero sum principle but in a seemingly different form. We need to identify the exact core problem solving pattern.

A little more thought immediately reveals to us the crucial technique of transforming the whole problem in the form of three variable zero sum principle.

We discover that we need to substitute for, $a^3 = p$, $-b^3 = q$ and $-c^3 = r$. This transforms the problem to the familiar form.

Problem transformation

With the above assumptions the problem is transformed now to,

If $p + q + r = 0$, then find, $p^3 + q^3 + r^3 - 3pqr$.

From the three variable zero sum principle we know the result to be zero.

Answer: Option a: $0$.

Key concepts used:

  • By End state analysis detecting a similarity to the results of familiar three variable zero sum principle.
  • Identifying the core problem solving pattern in the variables to be used for substitution.
  • Using substitution technique to transform the whole problem exactly in the form of three variable zero sum principle.

Final result was then obvious.

Problem 3.

If $a$, $b$ and $c$ are real and $a^2 + b^2 + c^2 = 2(a - b - c) - 3$, then $2a - 3b + 4c$ is,

  1. $0$
  2. $-1$
  3. $2$
  4. $1$

Problem analysis

While doing our End state analysis we observe, that the given expression is generally symmetric and regular in three variables, $a$, $b$ and $c$, but the target expression is completely assymetric in the three variables.

By our experience, we expect then to find the actual values of the variables from the given expression and know that only then we would be able to evaluate the target expression. 

With this sole objective we concentrate on the given expression and from the nature of the expression decide that the only way to find the values of three variables from a quadratic equation of three variables is to express the whole equation as a sum of squares equaling zero. That would give us the individual square terms equal to zeros and consequently reveal to us the values of all three variables in one shot.

This powerful principle is a much used Algebraic rich concept. We will call it as the Principle of zero sum of square tems.


Eighth Rich concept - Principle of zero sum of square terms

If we have an expression,

$p^2 + q^2 + r^2 = 0$, we conclude for real values of $p$, $q$ and $r$,

$p = q = r = 0$.

Problem transformation

With the goal of expressing the given expression as a sum of squares we start transforming it as,

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$,

Or, $(a^2 -2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$,

Or, $(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0$,

Or, $(a - 1) = (b + 1) = (c + 1) = 0$

Or, $a = 1$, $b = c = -1$, and,

$2a - 3b + 4c = 2 + 3 - 4 = 1$.

Answer: Option d: 1.

Key concepts used:

  • By end state analysis detecting an assymetry in the target expression hinting at the possibility of finding out the values of all three variables from the given expression directly at one shot.
  • Rearranging the terms of the given expression using Principle of collection of friendly terms, we form a sum of squares.
  • Using the principle of Principle of zero sum of square tems. we get each square term as zero and finally the values of each variable.

Eavaluating the final result was just a formality.


Here we have told you about the basic and rich concepts in Algebra, especially an extended set of powerful rich concepts that have been compiled from problem solving experiences.

Through analytical treatment of the solution process of a few selected sums it was shown how this basic and rich concept sets together with powerful general problem solving strategies enable elegant solution of the problems in a few simple steps.