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How to solve difficult Algebra problems in a few simple steps 3

How to Solve Difficult Algebra Problems Quickly in a few Steps 3

Difficult algebra problem solving techniques and Chosen problem solutions in very few steps

Chosen difficult algebra problems are solved quickly in very few steps. The problems are solved by no tricks, but easy algebra problem solving techniques.

Sections are,

  1. Causes behind difficulties in algebra problem solving
  2. Why algebra problem solving is interesting and how the problems are tackled
  3. Frequently used basic algebraic formulas (or algebraic identities): Needed in solving practically any hard algebra problem
  4. Advanced algebra problem solving concepts and techniques: For solving difficult algebra problems extra quick
  5. Quick solution of a few chosen difficult algebra problems: Problem solving techniques used in identifying key pattern for breakthrough and quick solution in very few steps explained clearly.

We'll repeat the usual reasons behind the difficulties faced by students in solving Algebra problems and the set of basic and rich concepts that are invaluable in reaching elegant solutions for seemingly difficult Algebra problems.

If you want, you may skip the first four sections and go straight to the problem solving stage. To skip click here.

You may directly move on to any of the last three sections in the above list by clicking on the link and return by clicking on browser back button.

Causes behind difficulties in Algebra problem solving

Following are the reasons for students finding algebra problems difficult,

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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 final performing stage. 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 algebra problem solving is interesting and how the problems are tackled

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 to common folks like us.

Concept layers

In our last article on Algebraic problem simplification, for the first time we have presented at length the concepts 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 new rich concepts are added to the layer incrementally. It is an extensible open ended layer of concepts.

We will repeat this time our basic concept layer as before for your convenience and more importantly will extend the rich concept layer by one more rich concept that we learned last time. This is incremental knowledge increase in action.

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

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

Basic Algebraic Concepts - Frequently used algebraic formulas

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 square of 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 $

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

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

Advanced algebra problem solving concepts often used in quick solution of difficult algebra problems

First rich algebra concept: 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.

You may refer to its effective use in more details in our second session on how to solve difficult Algebra problems in a few simple steps.

Second rich algebra concept: 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 pairwise 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 our first session on how to solve difficult Algebra problems in a few simple steps.

Third rich algebra concept: 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 rich algebra concept: 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 rich algebra concept: 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.

This highly useful technique has been used in our second session on how to solve difficult Algebra problems in a few simple steps as well as later in this session itself.

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

Sixth rich algebra concept: 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.

This great principle was again used effectively in our second session on how to solve difficult Algebra problems in a few simple steps.

Seventh rich algebra concept: 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.

You may refer to the effective use of this highly effective and essential technique in our second session on how to solve difficult Algebra problems in a few simple steps as well as later in the solution of a problem in this session itself.

Eighth rich algebra concept: Principle of zero sum of square terms

This principle is based on fundamental mathematical principles. It says,


$a^2 + b^2 + c^2 = 0$, and $a$, $b$ and $c$ are real, then, $a=b=c=0$.

On many occasions, use of this basic principle is the only way you can solve the problem quickly.

This powerful technique has been showcased while solving a problem in our second session on how to solve difficult Algebra problems in a few simple steps.

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

Quick solution of chosen difficult algebra problems in a few steps

Difficult algebra problem 1.

If $a^2 + 1 = a$, then the value of $a^{12} + a^6 + 1$ is,

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


First stage Problem analysis:

We notice first the large difference in powers of $a$ between the target and the given expressions.

Secondly, we don't find any apparent similarity between the two expressions. This is Principle of dissimilarity in action that resulted from End state analysis.

From prior experiences we have seen in such cases that the value of the variable can usually be derived from the given expression itself.

First stage input transformation

With this expectation, we focus our attention on the given expression and try to find ways of evaluating value of $a$ from the single quadratic equation itself.

$a^2 + 1 = a$,

Or, $a^2 - a + 1 = 0$.

First clue:

This is a simple transformation in itself, but the expression that we get reminds us of our sum of cubes expression and its factors specifically.

$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$.

It is up to us to recognize the useful pattern that the second factor of the sum of cubes expression corresponds exactly to our transformed given expression, where $x=a$ and $y=1$,

$a^3 + 1 = (a + 1)(a^2 - a + 1)$.

As this second factor is zero in pur problem, it will remain to be zero if we multiply it conveniently by $(a + 1)$, the first factor,

$a^3 + 1 = (a + 1)(a^2 - a + 1) = 0$, and so,

$a^3 + 1 = 0$,

Or, $a^3 = -1$.

Straightaway putting this value of $a^3$ in the target expression we have,

$a^{12} + a^6 + 1 = (a^3)^4 + (a^3)^2 + 1 = 1 + 1 + 1 = 3.$

Answer: Option c: $3$.

Key concepts used:

  • By the use of End State Analysis on the target end state expression, getting aware of the significant dissimilarity between the two expressions.
  • Forming an expectation of getting the value of the variable $a$ directly from the given expression itself.
  • Using input transformation technique as soon as we transform the given expression in a regular systematic form, we recognize it to be the second factor of the frequently used basic concept of sum of cubes expression. This is Pattern recognition skill in action resulting in use of Pattern recognition technique. Pattern recognition is one of the most important problem solving skills.
  • Now using the technique of using a new factor, we multiply the given expression by the first factor of sum of cubes expression.
  • From the resulting simple expression, we get the value of $a^3$ and use this value in the target expression directly by using Substitution technique.

Difficult algebra problem 2.

If $n^p - qn + \displaystyle\frac{1}{4}$, is to be a perfect square, the values of $p$ and $q$ would be,

  1. 2, 1
  2. 2, 3
  3. 2, -2
  4. 1, -1

Problem analysis and solution

To be a perfect square, the expression must be a quadratic expression, as in the middle term the power of $n$ is 1. It means, $p=2$.

By looking at the third term we find it to be square of $\displaystyle\frac{1}{2}$ and conclude that, if the expression is to be a perfect square, it must be of the form,

$\left(n - \displaystyle\frac{1}{2}\right)^2$.

So in the middle term, $2\times{\displaystyle\frac{1}{2}}n = qn$.

So, $q=1$.

Answer: Option a: 2, 1.

Key concepts used:

  • Using the concepts on the nature of terms of powers of sum of two variables, we conclude from the given expression that, $p=2$, that is, the expression is a quadratic one.
  • Now examining the third square term, we form the square expression itself.
  • At the last step, by the nature of middle term of a square of sum of two variables, we get the value of $q=1$.

This is a very simple problem, if you use your deductive reasoning based on the fundamental concepts on powers of sums of two variables.

We have included this problem here to highlight the importance of the basic concepts of algebra and ability to use the concepts.


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.

The list of Difficult algebra problem solving in a few steps quickly is available at, Quick algebra.

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