How to solve difficult Algebra problems in a few simple steps 5

How to solve difficult algebra problems in a few simple steps 5

Inverses, free resource and algebra concepts to solve problems in a few simple steps

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.

We'll not 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.

As a refresher on these, you may refer to our posts on basic and rich algebra concepts and Algebra problem simplification in a few steps 1, in a few steps 2, in a few steps 3 and in a few steps 4.

As principle of inverses plays an important role in solving the problems that we will take up here, we will repeat the powerful concept that is used so often to break the back of tough algebra problems.

Principle of inverses

This is one of the most useful concepts. You may refer to its detailed treatment in our article on principle of 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 positive integer of suitable value, we can always derive similar expressions in sum (or difference) 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, that is,

$x\times{\displaystyle\frac{1}{x}} = 1$.

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.

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

Problem 1.

If $a \gt 1$ and $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$, then the value of $a^4 - \displaystyle\frac{1}{a^4}$ is,

  1. $\frac{57895}{20736}$
  2. $\frac{57985}{20736}$
  3. $\frac{59825}{20736}$
  4. $\frac{58975}{20736}$


First stage Problem analysis:

We know from our experience with principle of inverses that with the given inverse expression in $a$ we should easily be able to get the value of sum of inverses of square of $a$.

Thinking till this point we stop and have a look at the target expression, comparing it with the result we have till now. This we will always do as a habit applying the end state analysis.

If we have the value of $a^2 + \displaystyle\frac{1}{a^2}$, to get the target expression $a^4 - \displaystyle\frac{1}{a^4}$ we need just to get the value of $a^2 - \displaystyle\frac{1}{a^2}$, so that the product gives the target expression value.

Then again by similar logic, we have $a + \displaystyle\frac{1}{a}$, and so to get $a^2 - \displaystyle\frac{1}{a^2}$, we need just to get $a - \displaystyle\frac{1}{a}$, so that the product gives the value of $a^2 - \displaystyle\frac{1}{a^2}$.

This is a clear path of deductive reasoning to decide on the shortest path to the solution. If you think deeper, you will find this reasoning to be a combination of end state analysis and working backwards approach.

What we need to do

So the analytical results provide us with two tasks, first to get $a^2 + \displaystyle\frac{1}{a^2}$ and then to get $a - \displaystyle\frac{1}{a}$, both from the given expression $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$.

Let's then get on with the job of implementing our decisions.

First stage input transformation to get $a^2 + \displaystyle\frac{1}{a^2}$

Knowing the use of principle of inverses it is easy for us,

$a + \displaystyle\frac{1}{a} = 2\frac{1}{12} = \frac{25}{12}$,

Or, $a^2 + \displaystyle\frac{1}{a^2} = \frac{625}{144} - 2 = \frac{337}{144}$

This calculation we had to do, and it has taken up a bit of our time, but still we are on target.

Second stage input transformation to get $a - \displaystyle\frac{1}{a}$

$a^2 + \displaystyle\frac{1}{a^2} = \displaystyle\frac{337}{144}$,

Or, $a^2 - 2 + \displaystyle\frac{1}{a^2} = \displaystyle\frac{337}{144} - 2$,

Or, $\left(a - \displaystyle\frac{1}{a}\right)^2 = \displaystyle\frac{49}{144}$,

Or, $\left(a - \displaystyle\frac{1}{a}\right) = \displaystyle\frac{7}{12}$, as $a \gt 1$.

By reuse technique, we have reused the value of sum of square of inverse, $a^2 + \displaystyle\frac{1}{a^2} = \displaystyle\frac{337}{144}$ to get the value of difference of inverse expression, $\left(a - \displaystyle\frac{1}{a}\right) = \displaystyle\frac{7}{12}$, directly without starting again from the given expression stage.

Final hurdle

We know $\left(a + \displaystyle\frac{1}{a}\right) = \displaystyle\frac{25}{12}$ and $\left(a - \displaystyle\frac{1}{a}\right) = \displaystyle\frac{7}{12}$.

So multiplying the two,

$\left(a^2 - \displaystyle\frac{1}{a^2}\right)=\displaystyle\frac{25\times{7}}{12^2}$.

Again we know,

$\left(a^2 + \displaystyle\frac{1}{a^2}\right)=\displaystyle\frac{337}{12^2}$.

Multiplying the two we get finally,

$\left(a^4 - \displaystyle\frac{1}{a^4}\right) = \displaystyle\frac{337\times{25}\times{7}}{12^4}$

Applying free resource use principle

In the beginning itself we have looked at the large choice values and decided to use whatever technique we know to avoid large multiplication. That is why we kept the factors, 7, 25 and 337 undisturbed without multiplying them.

First we test the numerators of choice values for factor 25 and quickly eliminate option a and b.

Next we test the two values of numerators of options c and d, that is, 59825 and 58975 for divisibility by 7. Either you apply the 7 divisibility technique or direct mental division here. We preferred the second approach and found 59825 not a multiple of 7 but 58975 to be a perfect multiple of 7. So the choice is option d.

Notice that we have ignored calculating $12^4$ altogether because all the denominators of four choices are same. If you actually calculate $12^4$, it will be equal to $20736$.

Answer: Option d: $\displaystyle\frac{58975}{20736}$.

Key concepts used:

  • By the use of End State Analysis comparing the target end state expression with the given expression, and knowing the potential of principle of inverses, we have used working backwards approach and basic algebra concepts further to decide on finding $\left(a^2 + \displaystyle\frac{1}{a^2}\right)$ first and then $\left(a - \displaystyle\frac{1}{a}\right)$.
  • After getting the value of $\left(a^2 + \displaystyle\frac{1}{a^2}\right)$, we have used the same value to get the value of $\left(a - \displaystyle\frac{1}{a}\right)$ without any cumbersome calculation.
  • In both cases we have used the input transformation technique.
  • Noticing the large choice values in the beginning itself we had decided to keep the factors of the products unevaluated till the last stage with the expectation of getting useful clue from the free resource of choice values. This delayed evaluation is a great technique as a time saving device in algebraic simplification often.
  • Using the basic concept of $(a + b)(a - b) = (a^2 - b^2)$ in two steps we reach the final stage.
  • Now we apply the free resource use principle and check the choice values quickly for divisibility of 7 and 25, thus isolating option d as the only possible answer.


This problem is not easy and even if you are able to solve, it might take up a bit if time in competitive test scenario. Our main recommendation is,

Go through the solution of the problem more than once and try other approaches to solution. It will be a great learning.

If you are not fluent in using suitable problem solving strategies along with pattern recognition skills, and basic and rich algebra concepts, you may prefer not to attempt this type of problem. You may just give a look at the large choice values, visualize the labor from start to end and skip.

But if you are very well practiced in using efficient problem solving strategies with basic and rich algebra concepts, using your pattern recognition skills and deductive reasoning, you may still decide to go ahead and solve it in about a minute's time.

Problem 2.

If $x = 3 + 2\sqrt{2}$, then the value of $\displaystyle\frac{x^6 + x^4 + x^2 + 1}{x^3}$ is,

  1. 192
  2. 216
  3. 204
  4. 198

Problem analysis and solution

Without waiting, we divide the numerator of the target expression by the denominator, because we notice the possibility of generating inverse expressions that we are comfortable with.

$\displaystyle\frac{x^6 + x^4 + x^2 + 1}{x^3}$

$= x^3 + x + \displaystyle\frac{1}{x} + \displaystyle\frac{1}{x^3}$

Our intention is to find a useful value of the inverse of $x$ and use that on the target expression.

Input transformation by Surd Rationalization Property

In dealing with surds, we examine a two term surd expression always to detect if it satisfies the Surd Rationalization Property which states,

If the difference of the square of the two terms in the surd expression is 1, when we inverse the expression and rationalize, its denominator will become 1, and the denominator will thus be eliminated very conveniently, leaving us with a second complementary surd expression.

For example, if the starting expression were $2 + \sqrt{3}$ that satisfies Surd rationalization property, after inverting and rationalizing we get as a result, its complementary expression, $2 - \sqrt{3}$.

Working in this line we now form the value of inverse of $x$ using rationalization technique,

$\displaystyle\frac{1}{x} = \displaystyle\frac{3 - \sqrt{2}}{3^2 - (2\sqrt{2})^2} = 3 - \sqrt{2}$, a very convenient expression.

With this result we get the value of sum of inverses,

$x + \displaystyle\frac{1}{x} = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$

Substituting this useful result, the target expression is transformed to,

$E = x^3 + \displaystyle\frac{1}{x^3} + 6$.

Knowing $x + \displaystyle\frac{1}{x} = 6$,

$x^2 + \displaystyle\frac{1}{x^2} = 6^2 - 2 = 34$, and

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

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

$=6\times{(34 - 1)} = 198$.

Finally then the value of the target expression is,

$E=x^3 + \displaystyle\frac{1}{x^3} + 6 = 204$.

Answer: Option c: 204.

Key concepts used:

  • Identifying the possible patterns of inverse expressions in the target expression and also identifying that inverse of given $x$ will be a useful value when transformed, we go for dividing the numerator of the target expression by the denominator and form the inverse expressions. This decision we reach by deductive reasoning that is holistic reasoning.
  • To form the sum of inverses, we have identified an important pattern in the given surd expression, and applied the concept of the Input tranformation by Surd rationalization property.
  • As expected, when we invert $x$, rationalize and then sum up $x$ and rationalized $\displaystyle\frac{1}{x}$, we get a very useful value of sum of inverse expression as a pure integer 6, the surd terms cancel out.
  • Reaching this point, last steps are easy and routine for us, as we are well versed now with technique of inverses.


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

It will be possible for you to reach the level of competence to solve any difficult problem within a minute with confidence. You need only to absorb the efficient problem solving concepts that are no tricks, and apply the concepts on actual problems a number of times along with clear knowledge of basic and rich algebra concepts.

Further Algebra resources

Exercise sessions on Algebra

SSC CGL Question set 1 on Algebra

SSC CGL Question set 8 on Algebra

SSC CGL Question set 9 on Algebra

SSC CGL Question set 10 on Algebra

SSC CGL Question set 11 on Algebra

SSC CGL Question set 13 on Algebra

Elegant solutions and explanations on Algebra question sets

SSC CGL Solution set 1 on Algebra

SSC CGL Solution set 8 on Algebra

SSC CGL Solution set 9 on Algebra

SSC CGL Solution set 10 on Algebra

SSC CGL Solution set 11 on Algebra

SSC CGL Solution set 13 on Algebra

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