Question

If a + b = 10 and ab = 21, then \[{{a}^{3}}+{{b}^{3}}=\]
(a) 270
(b) 370
(c) 470
(d) 170

Answer 1

Hint: First of all, write the formula for \[{{a}^{3}}+{{b}^{3}}\], that is \[\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]. Now here write \[\left( {{a}^{2}}+{{b}^{2}} \right)\] in terms of (a + b) and (ab) by using \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. Now, substitute the values of (a+b) and (ab) in the expression of \[{{a}^{3}}+{{b}^{3}}\] to finally get its value.
Complete step-by-step answer:
We are given that a + b = 10 and ab = 21. We have to find the value of \[{{a}^{3}}+{{b}^{3}}\]. Let us consider the expression given in the question.
\[E={{a}^{3}}+{{b}^{3}}\]
We know that,
\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]
By using this in the above expression, we get,
\[E=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]
We know that,
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab\]
By using this in the above expression, we get,
\[E=\left( a+b \right)\left[ \left( {{\left( a+b \right)}^{2}}-2ab \right)-ab \right]\]
\[E=\left( a+b \right)\left[ {{\left( a+b \right)}^{2}}-3ab \right]\]
Now, we are given that a + b = 10 and ab = 21. So, by substituting the value of (a + b) and ab in the above expression, we get,
\[E=\left( 10 \right)\left[ {{\left( 10 \right)}^{2}}-3\left( 21 \right) \right]\]
\[E=10\left[ 100-63 \right]\]
\[E=10\times 37\]
E = 370
So, we get the value of \[{{a}^{3}}+{{b}^{3}}\text{ as }370\].
Hence, option (b) is the right answer.

Note: In this question, many students try to find the values of a and b individually and then cube it but this method is not preferable as it can be very lengthy and can even lead to mistakes. Also, students can cross-check their answers by using the formula \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( ab+b \right)\]. By substituting the values of (a + b) = 10, ab = 21 and \[{{a}^{3}}+{{b}^{3}}=370\] in the above expression, we get,
\[{{\left( 10 \right)}^{3}}=370+3\left( 21 \right)\left( 10 \right)\]
\[1000=370+630\]
\[1000=1000\]
LHS = RHS
As we have got LHS = RHS. So, our answer is correct.