Question
If $a + b = 8$ and $ab = 6$, find the value of $a^3 + b^3.$
✓
Answer
We have, $a^3+b^3=(a+b)\left(a^2-a b+b^2\right)=(a+b)\left(a^2+b^2-a b\right)=(a+b)\left(a^2+b^2-a b+2 a b-2 a b\right)$
[Adding and substracting $2 a b$ in the second break $]
=(a+b)\left[\left(a^2+b^2+2 a b\right)-3 a b\right]=(a+b)\left[(a+b)^2-3 a b\right]\left[\because(a+b)^2=a^2+\right.$
$\left.b^2+2 a b\right]=8 \times\left[(8)^2-3 \times 6\right][\because a+b=8$ and $a b=6]=8 \times[64-18]=8 \times 46=368 \therefore a^3+b^3=368$
[Adding and substracting $2 a b$ in the second break $]
=(a+b)\left[\left(a^2+b^2+2 a b\right)-3 a b\right]=(a+b)\left[(a+b)^2-3 a b\right]\left[\because(a+b)^2=a^2+\right.$
$\left.b^2+2 a b\right]=8 \times\left[(8)^2-3 \times 6\right][\because a+b=8$ and $a b=6]=8 \times[64-18]=8 \times 46=368 \therefore a^3+b^3=368$
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