## Inverses, abstraction 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***and*

**in a few steps 3***.*

**in a few steps 4**You may watch the video of this article below.

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

**Solution:**

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

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 $x + \displaystyle\frac{1}{x} = 5$, then the value of $\displaystyle\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}$ is,

- $\displaystyle\frac{47}{21}$
- $\displaystyle\frac{41}{23}$
- $\displaystyle\frac{43}{23}$
- $\displaystyle\frac{45}{21}$

**Solution: ****First stage Problem analysis:**

As the given input is in the form of inverses, it is highly probable that we would have to use the techniques embodied in principle of inverses.

But looking at the **assymmetric fraction of large expressions** we decide not to evaluate inverse sums in other suitable powers of $x$.

Rather we focus our attention on the large fraction in the target expression.

**Simplification of this expression** is our first target.

**First stage - pattern identification**

With a closer look, we detect in no time that the numerator has an expression same as the denominator, that is, $x^4 +1$, though a bit hidden away.

We take this opportunity to immediately simplify. This is * simplification technique* at work - if you find an opportunity to simplify a complex expression, simplify it immediately without any delay.

Thus we have the target expression,

$E = 1 + \displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1} = 1 + \displaystyle\frac{E_1}{E_2}$, the fractional parts are given names.

This is a significant improvement even though the main barrier is not broken.

Again when we look at the numerator and compare it with the expanded form of the given expression. Using **Many ways technique****,** we get a hint of positive results.

As a first step we actually expand the given inverse expression. Perhaps for the first time we are doing this **instead of using the inherent powers of inverse expression.**

#### Second stage - applying Many ways technique

$x + \displaystyle\frac{1}{x} = 5$,

Or, $x^2 - 5x + 1 = 0$.

We will use this second form of input expression to simplify the target expression invoking Many ways technique.

Just as you try to solve a problem in many ways, you may need to use a given resource also in more than one way.

Here instead of using an inverse sum in its usual manner we are now using it in expanded second form.

#### Third stage - continued factor extraction technique for simplification

We have used this technique earlier and now we will use it again. When there is no other way, we apply this powerful algebraic technique for simplifying quite intimidating expressions with the help of a simpler input expression of zero value.

Essentially, we extract the input expression as a factor stage by stage from the higher order target expression and at each stage after extraction as a factor in a part of the target expression, we substitute zero value for the input expression to eliminate that part altogether.

Then again we take up second stage extraction of the input expression as a factor in a part of the remaining portion of the target expression.

At every stage the target expression thus gets simplified.

Let's see how this is done with the numerator of the fraction in the target expression.

$E_1 = 3x^3 + 5x^2 + 3x = 3x(x^2 - 5x + 1) + 15x^2 + 5x^2$

$= 20x^2$.

This is one stage factor extraction only but it shows the power of simplification in no uncertain terms.

So, the target expression is transformed to,

$E = 1 + \displaystyle\frac{20x^2}{x^4 + 1}$

$=1 + \displaystyle\frac{20}{x^2 + \displaystyle\frac{1}{x^2}}$

At last we have got our inverse expression in the target.

We evaluate the inverse in squares in no time,

$x + \displaystyle\frac{1}{x} = 5$,

Or, $x^2 + \displaystyle\frac{1}{x^2} = 25 -2 = 23$.

Thus we get,

$E = 1 + \displaystyle\frac{20}{23}$

$=\displaystyle\frac{43}{23}$

**Answer:** Option c: $\displaystyle\frac{43}{23}$.

**Comment:** Though it took a bit of time to explain, it finally has been a 40 seconds problem.

#### Key concepts used:

- By the use of
comparing the target end state expression with the given expression, we found no trace of inverse expression in the taget so decided on going through direct simplification.**End State Analysis** - As the numerator is the larger expression it was natural to focus on it. Examining it for a while first we have identified the presence of the denominator expression in the numerator expression and used
straightaway to simplify the target expression considerably.**Simplification technique** - Now again we give a close look on the numerator and get a hint of using the given expression in expanded form with equal to 0 on the RHS, as a means to simplify this large numerator.
- Using
we transform the input expression in the desired form of $x^2 - 5x + 1=0$. The idea is, wherever in the target expression we can extract the LHS of the equation, we will put 0 and eliminate that part of target expression. This is also use of**input transformation technique**as we are using an**Many ways technique,****inverse expression in a second way.** - Now only with a little bit of
**algebraic manipulation**, by dividing both numerator and denominator by $x^2$**we get our long awaited inverse sum in squares.** - Evaluating the
**value of inverse sum in squares from the given inverse sum was easy.** **Substitution of this value**gave us our final result.- The steps were few, but twists and turns were not so few.

### Problem 2.

If $3a^2 = b^2 \neq 0$, then the value of $\displaystyle\frac{(a + b)^3 - (a - b)^3}{(a + b)^2 + (a - b)^2}$ is,

- $\displaystyle\frac{b}{2}$
- $\displaystyle\frac{3b}{2}$
- $\displaystyle\frac{2b}{3}$
- $b$

**Problem analysis and solution**

Though the target expression is large we detect its symmetry in the form of presence of only two unique expressions $(a + b)$ and $(a - b)$. In such cases we use the not so commonplace form of *Substitution technique* using *Abstraction technique* by assuming $(a + b) = p$ and $(a - b) = q$ and substituting these in the target expression to transform it to a much simpler form.

**Note:** In the commonplace use of substitution technique we substitute a value of a simpler expression in a more complex expression to simplify it. We use this technique in algebra most frequently.

#### Target expression transformation using Abstraction technique and Substitution technique (second form)

Let's assume, $(a + b) = p$ and $(a - b) = q$ and substituting these in the target expression, we get,

$E = \displaystyle\frac{p^3 - q^3}{p^2 + q^2}$

Now we can easily recognize the familiar forms of basic relations in algebra,

$p^3 - q^3 = (p - q)(p^2 + pq + q^2)$.

Here in our case,

$p - q = a + b - a + b = 2b$,

$pq = (a + b)(a - b) = a^2 - b^2$, and

$p^2 + q^2 = (a + b)^2 + (a - b)^2 = 2a^2 + 2b^2$.

Using these values in the target expression,

$E = 2b\left(\displaystyle\frac{2a^2 + 2b^2 + a^2 - b^2}{2a^2 + 2b^2}\right)$

$=2b\left(\displaystyle\frac{3a^2 + b^2}{2a^2 + 2b^2}\right)$.

Now we use the given expression, $3a^2 = b^2 \neq 0$

$E = 2b\left(\displaystyle\frac{2b^2}{\displaystyle\frac{2b^2}{3} + 2b^2}\right)$

$=b\displaystyle\frac{2}{\displaystyle\frac{4}{3}}$

$=\displaystyle\frac{3b}{2}$

**Answer:** Option b: $\displaystyle\frac{3b}{2}$

#### Key concepts used:

- Abstracting $a + b$ and $a - b$ as $p$ and $q$ expressing the target in simplified form -- use of
and**Abstraction technique**.**Substitution technique** - Using the well known expression, $p^3 - q^3 = (p - q)(p^2 + pq + q^2)$ and evaluating all of the terms, $p^2 + q^2$, $p - q$ and $pq$.
- Substitution and simplification.

**Note:** Though this problem looked large, it was inherently simple. You know why? It is simple **because of the symmetry.**

### Summarization

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 is possible to reach the level of competence to solve any difficult algebra problem within a minute with confidence,** *if these concepts and techniques are first absorbed and then applied repeatedly through enough and systematic practice sessions devoted to solving selected not so easy problems. *

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

*.*

**Quick algebra**To go through the extended resource of **powerful concepts and methods** to solve difficult algebra problems, you may click on,