How to solve difficult SSC CGL algebra problems in a few steps 12

How to solve difficult SSC CGL algebra problems in a few steps 12

The powerful Continued factor extraction technique

Algebraic rich concepts and techniques along with general problem solving strategies enable elegant solution to the difficult SSL CGL Algebra problems for reaching the solution in quick time. The problem solving approach suggests first analysis of the problem followed by choice of strategy to deal with the problem. Goal is quick and sure solution.

In Algebra we have used many such powerful strategies and techniques for quick and elegant solution, but not always explained each in details.

In this session we will show how the important technique of Continued factor extraction can be applied for quick solution. We will use two specially selected problems for explanation.


Chosen Problem 1.

If $a+b+c=0$, then $\displaystyle\frac{a^2+b^2+c^2}{a^2-bc}$ is,

  1. 0
  2. 1
  3. 2
  4. 3

Problem analysis - first observation

The given equation with 0 on the RHS is a high value input information that is generally used at least in three ways to simplify the target expression,

  1. by transforming the equation in the forms of, $a+b=-c$, $b+c=-a$ or, $c+a=-b$, as needed,
  2. by taking out factor of $a+b+c$ from the target expression and replacing it by 0 for immediate significant simplification,
  3. by raising the three variable expression to square or cube and using the components of expanded expression for simplification.

On a quick analysis we couldn't find any of the above ways for use in simplifying the target expression easily.

Problem analysis - Second observation

Leaving this path of solution now we analyzed the target expression.

Many of the difficult algebra problems in SSC CGL involves simplification of a fraction with numerator and denominator. The answer choices being simple numerals here, we were sure that by some means the denominator will be completely eliminated, but it was not clear at this point how.

Problem analysis - Final choice of strategy

As we found no obvious way  for quick simplification, we decided to apply the never failing technique of Continued factor extraction on the target expression

Continued factor extraction technique

By this technique, at each step an additive factor of chosen factor expression is taken out from the more complex target expression by consuming one or more of the terms of the target expression and compensating for the other terms of the factor. The compensating terms are appended to the portion outside the factored part of the expression. The chosen factor expression is always selected in such a way that the factored part immediately reduces to 0 or similar such simplified value, leaving the rest of the target expression smaller and less complex. If solution is not reached by executing the first step of factor extraction, the same process is repeated till we reach our solution or see the path to the solution clearly.

Mark the crucial nature of the factor extraction process - we are extracting additive factors, not multiplicative, and by doing so we are reducing the complexity of the target expression at each step.

Let us apply the technique now to the target fraction expression to show how it actually works.

Problem solving execution - Choice of the factor and target expression

For applying the factor extraction technique, our target expression is the numerator of, $\displaystyle\frac{a^2+b^2+c^2}{a^2-bc}$, that is, $a^2+b^2+c^2$ and the chosen factor expression is the denominator $(a^2-bc)$ itself, because if we get an additive factor of this denominator expression in the numerator, the factored part will cease to be a part of the numerator by direct division and elimination of the denominator.

Almost always this technique can be applied to reduce the size and complexity of the more complex expression.

Note: $a^2 + b^2 + c^2$ is the target expression for immediate simplification only, otherwise the original fraction expression remains to be our main target for simplification.

First step of factor extraction

As we already have an $a^2$ common to the more complex numerator and the chosen factor denominator, we need just to subtract and add one term $bc$ to the numerator to take out one additive factor of $a^2-bc$,

$a^2+b^2+c^2=(a^2-bc) + bc + b^2 +c^2$.

In this step we have consumed $a^2$ in the factor and compensated $-bc$ in the factor by a $+bc$ outside the factor. Though this didn't reduce the size of the numerator, it eased the situation considerably because the numerator is transformed now from a three variable expression to a two variable expression.

So after the first step we have the target fraction in the form,

$E=\displaystyle\frac{(a^2-bc) + bc +b^2+c^2}{a^2-bc} = 1 + \displaystyle\frac{b^2+bc +c^2}{a^2-bc}$

Evaluation of the possibilities and final execution

As soon as we obtained a quadratic expression in $b$ and $c$, the possibility to use the input expression of $a+b+c=0$ in the form of $b+c=-a$ came into light. In this case it was easy to see through the key step as,

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

Or, $b^2 + bc + c^2 = a^2 - bc$.

Thus we get a second additive factor of $a^2 - bc$ from the numerator but this time by using the input expression giving the final result as 2.

This problem involves quick analysis and decision making based on the potential and nature of the possibilities and change of track as needed. Together the threads of reasoning we call deductive reasoning.

Answer: Option c : 2.

To show a more comprehensive application of the continued factor extraction technique, we will take up a second chosen problem. Here a chosen expression is extracted from the target step by step.

Chosen Problem 2.

If $\left(x + \displaystyle\frac{1}{x}\right)^2 = 3$ then the value of $(x^{72} + x^{66} + x^{54} + x^{36} + x^{24} + x^6 + 1)$ is,

  1. 4
  2. 2
  3. 3
  4. 1

Problem analysis:

We can visualize transforming the given expression first into a sum of square of inverses with a value of 1 and then proceed to try for further simplification. We must try for maximum simplification of the given expression, as the target expression is in very high powers of $x$.

First stage execution - extreme simplification of the input expression

$\left(x + \displaystyle\frac{1}{x}\right)^2 = 3$,

Or, $x^2 + 2 + \displaystyle\frac{1}{x^2} =3$

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

Or, $x^2 + \displaystyle\frac{1}{x^2} -1 = 0$.

Now we remember the sum of inverses in cube form as,

$x^3 + \displaystyle\frac{1}{x^3} =\left(x + \displaystyle\frac{1}{x}\right)\left(x^2 - 1 + \displaystyle\frac{1}{x^2}\right) = 0$.

This is a very significant simplification of the given expression and we know this will lead us to the solution.

Now our job is to find this factor of $x^3 + \displaystyle\frac{1}{x^3} =0$ step by step from the target expression in additive form and replace each such factored part by 0. This is direct application of Continued factor extraction technique.

Second stage execution - simplifying target expression using Continued factor extraction technique

The chosen factor expression is now,

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

Each time we extract the factor from the target expression, it is replaced by 0 and a part of target expression is eliminated straightway.

The target expression,

$E=(x^{72} + x^{66} + x^{54} + x^{36} + +x^{24} + x^6 + 1)$

$=x^{69}\left(x^3 + \displaystyle\frac{1}{x^3}\right) + x^{54} + x^{36} + x^{24} + x^6 + 1$

$=x^{54} + x^{36} + x^{24} + x^6 + 1$.

It is clear to us now, if we can form a pair of terms in powers of $x$ with difference in powers as 6, we can reduce the sum of the two terms to zero using simplified result of given expression.

We have achieved this simplification with the first two terms in highest powers of $x$. But as the difference between next two powers of $x$ is 18, we need to introduce two pairs of dummy terms with suitable powers,

$E=x^{54} + x^{36} + x^{24} + x^6 + 1$

$=x^{54} + x^{48} - x^{48} -  x^{42} + x^{42} + x^{36} + x^{24} + x^6 + 1$

$=x^{51}\left(x^3 + \displaystyle\frac{1}{x^3}\right) - x^{45}\left(x^3 + \displaystyle\frac{1}{x^3}\right) + x^{42} + x^{36} + x^{24} + x^6 + 1$

$=x^{42} + x^{36} + x^{24} + x^6 + 1$.

The first two terms results in to 0  and repeating the same process of introducing two pairs of dummy terms between $x^{24}$ and $x^6$,

$E=x^{24} + x^{18} - x^{18} - x^{12} + x^{12} + x^6 + 1$

$=1$, as three pairs of terms combine to 0.

Answer: Option d: 1.

Key concepts used: Simplifying the given sum of inverses to the maximum extent possible -- continued extraction of the simplified factor from the given expression.

This is a more difficult problem needing deeper problem solving abilities.

Note: Continued factor extraction technique uses the basic principle behind the process of division of algebraic expressions. Secondly, if you can use the method efficiently, seemingly intractable problems also can be solved very quickly. Thirdly, both these problems can be solved in a few tens of seconds wholly in your mind without putting pen on paper if you can reason and see through the steps, the abilities that can be acquired by intelligent practice of problem solving.