Question

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

Answer 1

Answer:-

Given:-

x + 2y + 3z = 0

• 2y + 3z = - x -- equation (1)

• x + 2y = - 3z -- equation (2)

• x + 3z = - 2y -- equation (3).

Also given that,

x³ + 4y³ + 9z³ = 18xyz

⟹ (x³ + 4y³ + 9z³)/xyz = 18 -- equation (4)

We have to find:-

$\implies \sf \: \dfrac{(x + 2y) ^{2} }{xy} + \dfrac{(2y + 3z) ^{2} }{yz} + \dfrac{(3z + x) ^{2} }{xz}$

Putting the respective values from equations (1) , (2) & (3) we get,

$\implies \sf \: \dfrac{( - 3z) ^{2} }{xy} + \dfrac{( - x) ^{2} }{yz} + \dfrac{( - 2y) ^{2} }{xz}$

Taking LCM we get,

$\implies \sf \: \dfrac{(yz)(xz)(9 {z}^{2} ) +(xy)(xz)(x ^{2} ) + (xy)(yz)( 4y ^{2} )}{xy \times yz \times xz} \\ \\ \\ \implies \sf \: \dfrac{ 9xy {z}^{4} + {x}^{4}yz + 4xy ^{4}z }{(xyz) ^{2} } \\ \\ \\ \implies \sf \: \frac{xyz( 9 {z}^{3} + {x}^{3} + 4y ^{3} ) }{(xyz)(xyz)} \\ \\ \\ \implies \sf \: \frac{ {x}^{3} + 4 {y}^{3} + 9 {z}^{3} }{xyz} \\ \\ \\ \implies \underline{ \underline{ \red{\sf \: 18}}} \: \: \: \: \: \sf \: [ \because \: from \: equation \: (4)]$

Answer 2

Answer:

Given :-

$\mapsto$ x + 2y + 3z = 0

$\mapsto$ x³ + 4y³ + 9z³ = 18xyz

To Evaluate :-

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

Solution :-

Given equation :

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

From this equation we get,

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

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

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

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

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

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

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

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

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

And,

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

Now,

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

By putting :

• x + 2y = - 3z
• 2y + 3z = - x
• 3z + x = - 2y

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

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

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

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

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

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

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

Here,

• x³ + 4y³ + 9z³ = 18xyz

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

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

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

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