Question

If 3x-2y=11 and xy=12, find the value of $27x^{3}-8y^{3}$.

Answer 1

Hint: In this question we have given 3x-2y=11 and xy=12, and we have to find the value of $27x^{3}-8y^{3}$, so for this we have to write this: $27x^{3}-8y^{3}$ as $\left( a^{3}-b^{3}\right) $ form and after that we have to use one formula, that is, $\left( a^{3}-b^{3}\right) =\left( a-b\right) \left( a^{2}+ab+b^{2}\right) $, by which we will get our required solution.

Complete step-by-step solution:
Given conditions are
3x-2y=11 …………………….equation(1)
xy=12 ………………………..equation(2)
Now, $27x^{3}-8y^{3}$ this can be written as,
$27x^{3}-8y^{3}$
$=3^{3}\times x^{3}-2^{3}\times y^{3}$
$=\left( 3x\right)^{3} -\left( 2y\right)^{3} $ [since, as we know that $a^{n}\times b^{n}=\left( ab\right)^{n} $]
Now by using $\left( a^{3}-b^{3}\right) =\left( a-b\right) \left( a^{2}+ab+b^{2}\right) $..........equation(3)
We get,
$=\left( 3x\right)^{3} -\left( 2y\right)^{3} $ [ we have used a=3x and b=2y]
$=\left( 3x-2y\right) \left\{ \left( 3x\right)^{2} +\left( 3x\right) \left( 2y\right) +\left( 2y\right)^{2} \right\} $
$=\left( 3x-2y\right) \left\{ \left( 3x\right)^{2} +6xy+\left( 2y\right)^{2} \right\} $
$=11\times \left\{ \left( 3x\right)^{2} +6\times 12+\left( 2y\right)^{2} \right\} $ [using equation(1) and (2)]
$=11\times \left\{ \left( 3x\right)^{2} +\left( 2y\right)^{2} +72\right\} $ ………...equation(4)

So to find the solution of the above equation, we have to find the value of $\left( 3x\right)^{2} +\left( 2y\right)^{2} $, so for this we have to use identity , that is, $ \left( a\right)^{2} +\left( b\right)^{2} =\left( a-b\right)^{2} +2ab$
So by the above formula we can write $\left( 3x\right)^{2} +\left( 2y\right)^{2} $ as,
$\left\{ \left( 3x-2y\right)^{2} +2\left( 3x\right) \left( 2y\right) \right\} $
=$\left\{ \left( 3x-2y\right)^{2} +12xy\right\} $
=$\left\{ \left( 11\right)^{2} +12\times 12\right\} $ [using equation (1) and (2)]
=121+144=265.

So therefore, we get $\left( 3x\right)^{2} +\left( 2y\right)^{2} $ =265.
Now putting the value of $\left( 3x\right)^{2} +\left( 2y\right)^{2} $ in equation(4), we get,
$11\times \left\{ \left( 3x\right)^{2} +\left( 2y\right)^{2} +72\right\} $
=$11\times \left\{ 265+72\right\} $
=$11\times 337$=3707.

Therefore, we can write $27x^{3}-8y^{3}$=3707, which is our required solution.

Note: To solve this type of problems we need to keep in mind those formulas that we have used in this solution, also you might get confused that why we separately find the value of $\left( 3x\right)^{2} +\left( 2y\right)^{2} $, because in this question we have given only the values of 3x-2y=11 and xy=12 and there is no direct value of $\left( 3x\right)^{2} +\left( 2y\right)^{2} $ , so we have to use these( i.e,3x-2y=11 and xy=12) to find its value.