Question

Solve the following equations:
$3x^3-8x{y}^2+y^3+21=0,x^2(y-x)=1$

• A. $1,2$
• B. $-1,3$
• C. $,\sqrt[3]{3\frac{1}{2}},3\sqrt[3]{3\frac{1}{2}}$.
• D. $\sqrt{\frac{3}{2}},\frac{1}{2}$

Answer 1

The correct option is C $,\sqrt[3]{3\frac{1}{2}},3\sqrt[3]{3\frac{1}{2}}$.

Given, $x^2(y-x)=1$

$\Rightarrow x^{2}y-x^3=1y=\frac{1+x^3}{x^2}$ ......(i)

Also given, $3x^3-8x^{2}y+y^3+21=0$

Substituting $y$ from (i) in above equation,

$3x^3-8x^2\left(\frac{1+x^3}{x^2}\right)+{\left(\frac{1+x^3}{x^2}\right)}^3+21=0\Rightarrow -4x^9+8x^6-5x^3+1=0\Rightarrow 4x^9-8x^6+5x^3-1=0\Rightarrow 4x^9-4x^6-4x^6+4x^3+x^3-1=0\Rightarrow 4x^6(x^3-1)-4x^3(x^3-1)+1(x^3-1)=0\Rightarrow (x^3-1)(4x^6-4x^3+1)=0\Rightarrow (x^3-1)=0\Rightarrow x=1$

`Also $4x^6-4x^3+1=0$

$\Rightarrow {(2x^3-1)}^2=0\Rightarrow 2x^3-1=0\Rightarrow x=\sqrt[3]{3\frac{1}{2}}$

Substituting $x$ in (i), we get

For $x=1$

$\Rightarrow y=\frac{1+1}{1}=2$

For $x=\sqrt[3]{3\frac{1}{2}}$

$\Rightarrow y=\frac{1+\frac{1}{2}}{{\left(\frac{1}{2}\right)}^{\frac{2}{3}}}=3\sqrt[3]{3\frac{1}{2}}$

So, the values of $x$ are $1,\sqrt[3]{3\frac{1}{2}}$ and values of $y$ are $\frac{1}{2},3\sqrt[3]{3\frac{1}{2}}$.