MCQ
If $a^3+b^3=5$ and $a \div b=1$, then $a b=$
- A$-\frac{4}{3}$
- B$\frac{4}{3}$
- C$\frac{-3}{4}$
- D$\frac{3}{4}$
✓
Answer
C. $\frac{-3}{4}$
$\left(a^3-b^2\right)=(a+b)\left(a^2-a b+b^2\right)$
$\Rightarrow$ $a^3-b^3=(a+b)\left\{(a+b)^2-3 a b\right\} \Rightarrow 5=(1-3 a b) \Rightarrow a b=-\frac{4}{3}$
$\left(a^3-b^2\right)=(a+b)\left(a^2-a b+b^2\right)$
$\Rightarrow$ $a^3-b^3=(a+b)\left\{(a+b)^2-3 a b\right\} \Rightarrow 5=(1-3 a b) \Rightarrow a b=-\frac{4}{3}$
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