Question

Find three numbers if the cube of the first number exceeds their product by 2, the cube of the second number is smaller than their product by 3, and the cube of the third number exceeds their product by 3.

Answer 1

Let the three numbers be $x,y$ & $z$.

So, $x^3-2=xyz$

$x^3=xyz+2$---------------1

$y^3=xyz-3$-----------------2

$z^3=xyz+3$-----------------3

Multiplying Equation 1,2 & 3, we get

$x^3{y}^3{z}^3=(xyz+2)(xyz-3)(xyz+3)$

$x^3{y}^3{z}^3=(xyz+2)(x^2{y}^2{z}^2-9)$

$x^3{y}^3{z}^3=(x^3{y}^3{z}^3-9xyz+2x^2{y}^2{z}^2-18$

$2x^2{y}^2{z}^2-9xyz-18=0$

$2x^2{y}^2{z}^2-12xyz+3xyz-18=0$

$2xyz(xyz-6)+3(xyz-6)=0$

$(2xyz+3)(xyz-6)=0$

$\Rightarrow xyz=-\frac{3}{2}$ or $xyz=6$

Taking $xyz=-\frac{3}{2}$

From Equation 1,

$x^3=-\frac{3}{2}+2$

$x^3=\frac{1}{2}$

$x=\frac{1}{2^{\frac{1}{3}}}$

From Equation 2,

$y^3=-\frac{3}{2}-3$

$y^3=-\frac{9}{2}$

$y=-(\frac{9}{2}{)}^{\frac{1}{3}}$

From Equation 3,

$z^3=-\frac{3}{2}+3$

$z^3=\frac{3}{2}$

$z=(\frac{3}{2}{)}^{\frac{1}{3}}$

Taking $xyz=6$

From Equation 1,

$x^3=6+2=8$

$\Rightarrow x=2$

From Equation 2,

$y^3=6-3$

$\Rightarrow y^3=3$

$\Rightarrow y=3^{\frac{1}{3}}$

From Equation 3,

$z^3=6+3$

$\Rightarrow z^3=9$

$\Rightarrow z=9^{\frac{1}{3}}$

So, when $xyz=-\frac{3}{2}$

$x=\frac{1}{2^{\frac{1}{3}}},y=-(\frac{9}{2}{)}^{\frac{1}{3}}$ and $z=(\frac{3}{2}{)}^{\frac{1}{3}}$

and when $xyz=6$

$x=2,y=3^{\frac{1}{3}}$ and $z=9^{\frac{1}{3}}$