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 ab in the second break]
$=(a+b)\left[\left(a^2+b^2+2 a b\right)-3 a b\right]$
$=(a+b)\left[(a+b)^2-3 a b\right]$
${\left[\because(a+b)^2=a^2+b^2+2 a b\right]}$
$=8 \times\left[(8)^2-3 \times 6\right][\because a+b=8 \text { and } a b=6]$
$=8 \times[64-18]$
$=8 \times 46$
$=368$
$\therefore a^3+b^3=368$
$a^3+b^3=(a+b)\left(a^2-a b+b^2\right)$
$=(a+b)\left(a^2+b^2-a b\right)$
$=(a+b)\left(a^2+b^2-a b+2 a b-2 a b\right)$
[Adding and substracting 2 ab in the second break]
$=(a+b)\left[\left(a^2+b^2+2 a b\right)-3 a b\right]$
$=(a+b)\left[(a+b)^2-3 a b\right]$
${\left[\because(a+b)^2=a^2+b^2+2 a b\right]}$
$=8 \times\left[(8)^2-3 \times 6\right][\because a+b=8 \text { and } a b=6]$
$=8 \times[64-18]$
$=8 \times 46$
$=368$
$\therefore a^3+b^3=368$
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