Question

Evaluate the following:
\[64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}\]

Answer 1

Hint: In this question, first of all, rearrange the convert the terms of the given expression in the form of \[{{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\]. Now, we know that \[{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\]. So, convert the expression accordingly.

Complete step-by-step answer:

In this question, we need to factorize the expression,

\[64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}\]

Let us consider the expression given in the question.

\[E=64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}\]

We can also write the above expression as:

\[E={{\left( 4 \right)}^{3}}.{{a}^{3}}-{{\left( 3 \right)}^{3}}.{{b}^{3}}-3\times \left( 16 \right).3{{a}^{2}}.b+3\times 4\times \left( 9 \right)a.{{b}^{2}}\]

We know that \[{{a}^{m}}.{{b}^{m}}={{\left( a.b \right)}^{m}}\]. By using this in the above expression, we get

\[E={{\left( 4a \right)}^{3}}-{{\left( 3b \right)}^{3}}-3\times {{\left( 4 \right)}^{2}}.{{a}^{2}}.3.b+3.{{\left( 3 \right)}^{2}}.{{b}^{2}}.4a\]

So, we can also write the above expression as:

\[E={{\left( 4a \right)}^{3}}-{{\left( 3b \right)}^{3}}-3{{\left( 4a \right)}^{2}}.\left( 3b \right)+3{{\left( 3b \right)}^{2}}.4a\]

By taking out 3.(4a)(3b) common from the last two terms of the above equation, we get,

\[E={{\left( 4a \right)}^{3}}-{{\left( 3b \right)}^{3}}-3.4a.3b\left[ \left( 4a \right)-\left( 3b \right) \right]\]

By substituting 4a = A and 3b = B in the above expression, we get,

\[E={{A}^{3}}-{{B}^{3}}-3A.B\left( A-B \right)\]

We know that \[\left[ {{m}^{3}}-{{n}^{3}}-3mn\left( m-n \right) \right]={{\left( m-n \right)}^{3}}\]. By using this in the above expression, we get,

\[E={{\left( A-B \right)}^{3}}\]

Hence, we get,

\[64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}={{\left( 4a-3b \right)}^{3}}\]


Note: In this question, students can verify the answer by expanding the expression we have got as answers and checking if it is equal to the expression given in the question or not in the following way.

We have got

\[E={{\left( 4a-3b \right)}^{3}}\]

We can also write the above expression as:

\[E=\left( 4a-3b \right)\left( 4a-3b \right)\left( 4a-3b \right)\]

Now by multiplying the above terms by simple algebraic rules, we get,

\[E=\left[ \left( 4a.4a \right)-\left( 4a \right).\left( 3b \right)-\left( 3b \right).\left( 4a \right)+\left( 3b \right)\left( 3b \right) \right]\left( 4a-3b \right)\]

\[E=\left( 16{{a}^{2}}-12ba-12ba+9{{b}^{2}} \right)\left( 4a-3b \right)\]

\[E=\left( 16{{a}^{2}}-24ab+9{{b}^{2}} \right)\left( 4a-3b \right)\]

By further simplifying the above expression, we get,

\[E=64{{a}^{3}}-48{{a}^{2}}b-96{{a}^{2}}b+36a{{b}^{2}}+72a{{b}^{2}}-27{{b}^{3}}\]

\[E=64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}\]

Hence, this expression is equal to the original expression. Therefore, our answer is correct.