Math, asked by AbhirVishwa0233, 9 months ago

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

Answers

Answered by Anonymous
22

\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

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