Question

If $2x+3y=7$ and $xy=2,$ find the value of $8x^3+27y^3$

Answer 1

Given: $2x+3y=7$ $...\left(1\right)$

and $xy=2$ $...\left(2\right)$

$\Rightarrow y=\frac{2}{x}$

Substituting value of $y$ in $\left(1\right),$

$\Rightarrow 2x+3\times \frac{2}{x}=7$

$\Rightarrow 2x^2-7x+6=0$

$\Rightarrow 2x^2-4x-3x+6=0$

$\Rightarrow 2x(x-2)-3(x-2)=0$

$\Rightarrow (2x-3)(x-2)=0$

$\Rightarrow x=\frac{3}{2}$ or $x=2$

$\therefore y=\frac{2}{2}=1$

$\therefore y=\frac{2}{\frac{3}{2}}=\frac{4}{3}$

Possible values of $(x,y)=(\frac{3}{2},\frac{4}{3}),(2,1)$

If $(x,y)=(\frac{3}{2},\frac{4}{3})$ then value of $8x^3+27y^3$

$=8\times {\left(\frac{3}{2}\right)}^3+27{\left(\frac{4}{3}\right)}^3$

$=8\times \frac{27}{8}+27\times \frac{64}{27}$

$=27+64$

$=91$

If $(x,y)=(2,1)$ then value of $8x^3+27y^3$

$=8\times (2{)}^3+27$

$=8\times 8+27$

$=64+27$

$=91$