Question

If p= ( 2-a), then the value of ${{a}^{3}}+6ap+{{p}^{3}}-8$ is.
a. 0
b. 1
c. 2
d. 3

Answer 1

Hint: Here, we have been given relation between p and a, that is ${{a}^{3}}+6ap+{{p}^{3}}-8$ and one more equation, that is, $p=2-a$ is given to us. And to find the value of the given equation in terms of p and a, we will substitute $p=2-a$ or $a=2-p$ and put it in the equation, ${{a}^{3}}+6ap+{{p}^{3}}-8$ and solve to find the answer.

Complete step-by-step solution:
We have been given in the question, a general equation with two variables a and p, and a simple equation between a and p, that is, $p=2-a$. We can proceed by substituting the value of $p=2-a$ in the equation that has been given to us in the question, which is, ${{a}^{3}}+6ap+{{p}^{3}}-8$.
So, when we substitute, we get as follows.
${{a}^{3}}+6a\left( 2-a \right)+{{\left( 2-a \right)}^{3}}-8$
We know that, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}$. So, by using this identity, we can write the above equation as follows.
$\begin{align}
  & {{a}^{3}}+12a-6{{a}^{2}}+8-{{a}^{3}}-3\times 4a+3\times 2{{a}^{2}}-8 \\
 & ={{a}^{3}}+12a-6{{a}^{2}}+8-{{a}^{3}}-12a+6{{a}^{2}}-8 \\
\end{align}$
On grouping the similar terms, we get,
$={{a}^{3}}-{{a}^{3}}+12a-12a-6{{a}^{2}}+6{{a}^{2}}+8-8$
We can see each term gets cancelled out. Therefore, we get,
= 0
Hence, we get the value of the expression as 0.

Note: Here, as we know that we have substituted the value $p=2-a$ in the equation ${{a}^{3}}+6ap+{{p}^{3}}-8$. We can also substitute the value of a, as $a=2-p$ in the equation, ${{a}^{3}}+6ap+{{p}^{3}}-8$. So, we will get,
${{\left( 2-p \right)}^{3}}+6\left( 2-p \right)p+{{p}^{3}}-8$
And on using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}$, we will get as,
$\begin{align}
  & {{2}^{3}}-{{p}^{3}}-3\times {{2}^{2}}p+3\times 2{{p}^{2}}+12p-6{{p}^{2}}+{{p}^{3}}-8 \\
 & =8-{{p}^{3}}-12p+6{{p}^{2}}+12p-6{{p}^{2}}+{{p}^{3}}-8 \\
 & =0 \\
\end{align}$
So, by following this method also, we get the answer as 0.
Also, the students must ensure that they write the expansion of the identity, ${{\left( x-y \right)}^{3}}$ correctly, as some students may make mistake in it and may write it as ${{\left( x-y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y-3x{{y}^{2}}$, but the correct formula is ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}$.