Question

If $a+b+c=9,{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=35$, then the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ is
[a] 100
[b] 108
[c] 95
[d] 1008

Answer 1

Hint: Square both sides of the equation a+b+c = 9 and use the fact that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$. Substitute the value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$ and hence find the value of $ab+bc+ac$. Use the fact that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$. Substitute the values of $a+b+c,{{a}^{2}}+{{b}^{2}}+{{c}^{2}},ab+bc+ac$ and hence find the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$. Hence determine which of the options are correct.

Complete step by step answer:
We have $a+b+c=9\ \ \left( i \right)$
Squaring both sides, we get
${{\left( a+b+c \right)}^{2}}={{9}^{2}}=81$
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$
Using the above identity, we get
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac=81$
Given that ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=35\text{ }\left( ii \right)$. Substituting the value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$, we get
$35+2ab+2bc+2ac=81$
Subtracting 35 from both sides, we get
$2ab+2bc+2ac=46$
Dividing both sides by 2, we get
$ab+bc+ac=23\text{ }\left( iii \right)$
We know that
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$
Taking -1 common from the last three terms of the second factor on RHS, we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-\left( ab+bc+ac \right) \right)$
From equation (i), equation (ii), and equation (iii), we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( 9 \right)\left( 35-23 \right)=108$

So, the correct answer is “Option B”.

Note: [1] Alternative solution: We can solve the question using the fact that the cubic equation whose sum of roots is a, product of roots taken two at a time is b and product of roots taken all at a time is c is given by
${{x}^{3}}-a{{x}^{2}}+bx-c=0$
Let $abc=t$
Hence, we have from equation (i) and equation (ii), the cubic equation whose roots are given by a, b and c is given by
${{x}^{3}}-9{{x}^{2}}+23x-t=0$
Adding $9{{x}^{2}}-23x+t$ on both sides, we get
${{x}^{3}}=9{{x}^{2}}-23x+t$
Since a, b, c satisfy this equation, we have
$\begin{align}
  & {{a}^{3}}=9{{a}^{2}}-23a+t\ \ \left( iv \right) \\
 & {{b}^{3}}=9{{b}^{2}}-23b+t\ \ \ \left( v \right) \\
 & {{c}^{3}}\ =9{{c}^{2}}-23c+t\ \ \left( vi \right) \\
\end{align}$
Adding equation (iv), equation (v) and equation (vi), we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=9\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)-23\left( a+b+c \right)+3t$
From equation (i) and equation (ii), we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=9\times 35-23\times 9+3t$
Subtracting 3t from both sides, we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3t=108$
Reverting to original variable, we get
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=108$, which is the same as obtained above.
Hence option [b] is correct.