If x3-4 y3-9 y3-18 x y z and x-2 y-3 z-0- evaluatex-2 y2-x y-2 y-3 z2-y z-3 z-x2-2 x

The correct option is A 18Given that x3 + 4y3 + 9z3 = 18xyz and x + 2y + 3z = 0 ∴ x + 2y = 3z, 2y + 3z = x and 3z + x = 2y (x+2y)^2/xy + (2y + 3z)^2/yz + ...

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Byju's Answer

Standard IX

Mathematics

Expansion of (a ± b)³

If x3+4 y3+9 ...

Question

If x³ + 4y³ + 9z³ = 18xyz and x + 2y + 3z = 0 ; evaluate
$\frac{(x + 2 y)^{2}}{x y} + \frac{(2 y + 3 z)^{2}}{y z} + \frac{(3 z + x)^{2}}{z x}$

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Solution

The correct option is A 18
Given that x³ + 4y³ + 9z³ = 18xyz and x + 2y + 3z = 0
$∴$ x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y

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

= $\frac{9 z^{2}}{x y} + \frac{x^{2}}{y z} + \frac{4 y^{2}}{z x}$

= $\frac{x^{3} + 4 y^{3} + 9 z^{3}}{x y z}$

Given that x³ + 4y³ + 9z³ = 18xyz

$∴ \frac{(x + 2 y)^{2}}{x y} + \frac{(2 y + 3 z)^{2}}{y z} + \frac{(3 z + x)^{2}}{z x} = \frac{18 x y z}{x y z}$

= 18

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Similar questions

Q. If

x + 2 y + 3 z = 0

and

x^{3} + 4 y^{3} + 9 z^{3} = 18 x y z;

evaluate :

\frac{(x + 2 y)^{2}}{x y} + \frac{(2 y + 3 z)^{2}}{y z} + \frac{(3 z + x)^{2}}{z x}

Q. State True or False.

x^{4} - (2 y - 3 z)^{2} =

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

If $x + 2 y + 3 z = 0$ then $x^{3} + 8 y^{3} + 27 z^{3}$ = _____________

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I.

x^{2} + y^{2} + Z^{2} = 0

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

If x³ + 4y³ + 9z³ = 18xyz and x + 2y + 3z = 0 ; evaluate
$\frac{(x + 2 y)^{2}}{x y} + \frac{(2 y + 3 z)^{2}}{y z} + \frac{(3 z + x)^{2}}{z x}$