Question

Solve the following systems of equations.
$\sqrt[3]{3x}+\sqrt[3]{3y}=4,x+y=28$

Answer 1

$(x{)}^{1/3}+(y{)}^{1/3}=4,x+y=28$

$x^{1/3}+y^{1/3}=4$

Cubiy both sides.

$(x+y)+3.\left(xy{)}^{1/3}\right(x^{1/3}+y^{1/3})=6y$

$\Rightarrow 28+3.\left(xy{)}^{1/3}\right(4)=6y$

$12.(xy{)}^{1/3}=36$

$(xy{)}^{1/3}=3$

$xy=27$

Using

$(x-y{)}^2=(x+y{)}^2-4xy$

$=(28{)}^2-4.(27)$

$=784-108$

$(x-y{)}^2=676$

$x-y=\pm 26$

$x+y=28$

__________

$2x=5y$ or $2$

$x=27$ or $1$

When $x=27,y=1$

$x=1,y=27$