Question

if x+2y+3z=0, then show that x^3+4y^3+27z^3=18xyz

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★ $x+2y+3z = 0$

$\huge\tt{\red{\underline{To\:\:Prove:}}}$

★$x^{3}+8y^{3}+27z^{3}=18xyz$

(correct statement to be proved)

$\huge\tt{\red{\underline{Proof:}}}$

We have,

$x+2y+3z = 0$

& $x^{3}+8y^{3}+27z^{3}=x^{3}+(2y)^{3}+(3z)^{3}$

Let us consider

•x as a

•2y as b

•3z as c.

Then,

$a+b+c=0$

&

$x^{3}+8y^{3}+27z^{3}=a^{3}+b^{3}+c^{3}=18xyz$

We know,

We know, if a + b + c = 0

Then, $a^{3}+b^{3}+c^{3}=3abc$

. °. $x^{3}+8y^{3}+27z^{3}$

=$3\times x \times 3y \times 2z$

=$18xyz$

Hence proved