Question

1. If x + 2y + 3z = 0 and
x3 + 4y3 + 9z3 = 18xyz; evaluate :
(x + 2y)2 (2y+3z)? (3z + x)2
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Answer 1

Step-by-step explanation:

11

ANSWER

Given, x+2y+3z=0  and   x3+4y3+9z3=18xyz

      =>x+2y=−3z

also, =>2y+3z=−x

and   =>3z+x=−2y

Now,

xy(x+2y)2+yz(2y+3z)2+zx(3z+x)2

=xy(−3z)2+yz(−x)2+zx(−2y)2

=xy9z2+zyx2+zx4y2

=xyz9z3+xyzx3+xyz4y3

=xyz9z3+x3+4y3

=xyz18xyz

=18

Answer 2

${\huge{\boxed{\sf{\green{❥✰Question✰}}}}}$

If x + 2y + 3z = 0 and

x3 + 4y3 + 9z3 = 18xyz; evaluate :

(x+2y)2/xy+(2y + 3z)2/yz+(3z + x)2/zx ?

$\huge\mathbb\fcolorbox{Green}{violet}{♡ᎪղՏωᎬя᭄}$

$\mapsto$  x + 2y + 3z = 0

$\mapsto$ x³ + 4y³ + 9z³ = 18xyz

To Evaluate :-

$\begin{gathered}\mapsto \sf \dfrac{{(x + 2y)}^{2}}{xy} + \dfrac{{(2y + 3z)}^{2}}{yz} + \dfrac{{(3z + x)}^{2}}{zx}\\\end{gathered}$

Solution :-

Given equation :

$\Rightarrow \sf x + 2y + 3z =\: 0⇒x+2y+3z=0$

From this equation we get,

${\small{\bold{\purple{\underline{\bigstar\: In\: the\: {1}^{{st}}\: case\: :-}}}}}$

$\implies \sf x + 2y + 3z =\: 0⟹x+2y+3z=0$

$\begin{gathered}\implies \sf \bold{\green{x + 2y =\: - 3z\: ------\: (Equation\: No\: 1)}}\\\end{gathered}$

${\small{\bold{\purple{\underline{\bigstar\: In\: the\: {2}^{{nd}}\: case\: :-}}}}}$

$\implies \sf x + 2y + 3z =\: 0⟹x+2y+3z=0$

$\implies \sf x + 2y + 3z =\: 0⟹x+2y+3z=0$

$\begin{gathered}\implies \sf \bold{\green{2y + 3z =\: - x\: ------\: (Equation\: No\: 2)}}\\\end{gathered}$

${\small{\bold{\purple{\underline{\bigstar\: In\: the\: {3}^{{rd}}\: case\: :-}}}}}$

$\implies \sf x + 2y + 3z =\: 0⟹x+2y+3z=0$

$\begin{gathered}\implies \sf\bold{\green{3z + x=\: - 2y\: ------\: (Equation\: No\: 3)}}\\\end{gathered}$

And,

$\Rightarrow \sf x^3 + 4y^3 + 9z^3 =\: 18xyz⇒x$

Now,

$\begin{gathered}\dashrightarrow \sf \dfrac{{(x + 2y)}^{2}}{xy} + \dfrac{{(2y + 3z)}^{2}}{yz} + \dfrac{{(3z + x)}^{2}}{zx}\\\end{gathered}$

By putting :

$\begin{gathered}\dashrightarrow \sf \dfrac{{(- 3z)}^{2}}{xy} + \dfrac{{(- x)}^{2}}{yz} + \dfrac{{(- 2y)}^{2}}{zx}\\\end{gathered}$

$\begin{gathered}\dashrightarrow \sf \dfrac{9z^2}{xy} + \dfrac{x^2}{yz} + \dfrac{4y^2}{zx}\\\end{gathered}$

$\begin{gathered}\dashrightarrow \sf \dfrac{(yz)(xz)(9z^2) + (xy)(xz)(x^2) + (xy)(yz)(4y^2)}{xy \times yz \times xz}\\\end{gathered}$

$\begin{gathered}\dashrightarrow \sf \dfrac{(9xyz^4) + (x^4 yz) + 4xy^4z)}{(xyz)^2}\\\end{gathered}$

$\begin{gathered}\dashrightarrow \sf \dfrac{\cancel{xyz}(9z^3 + x^3 + 4y^3)}{\cancel{(xyz)} \times (xyz)}\\\end{gathered}$

$\begin{gathered}\dashrightarrow \sf \dfrac{9z^3 + x^3 + 4y^3}{xyz}\\\end{gathered}$

$\dashrightarrow \sf \dfrac{x^3 + 4y^3 + 9z^3}{xyz}⇢$

Here,

$\dashrightarrow \sf \dfrac{18\cancel{xyz}}{\cancel{xyz}}$

$\dashrightarrow \sf \dfrac{18}{1}$

$\dashrightarrow \sf\boxed{\bold{\red{18}}}⇢$

$\begin{gathered}\therefore \sf \dfrac{{(x + 2y)}^{2}}{xy} + \dfrac{{(2y + 3z)}^{2}}{yz} + \dfrac{{(3z + x)}^{2}}{zx} =\: \bold{\red{18}}.\\\end{gathered}$

Hence the answer us 18