If ${(x + y)}^{3} - z^{2} = 4$ , ${(y + z)}^{2} - x^{2} = 9$ and ${(z + x)}^{2} - y^{2} = 36$ , then find the value of $x + y + z$

7

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $7$
${(x + y)}^{2} - z^{2} = 4 \Rightarrow (x + y + z) (x + y - z) = 4$
$\Rightarrow (x + y - z) = \frac{4}{x + y + z} . . . . (i)$
${(y + z)}^{2} - x^{2} = 9 \Rightarrow (y + z + x) (y + z - x) = 9$
$\Rightarrow (y + z - x) = \frac{9}{x + y + z} . . . . (i i)$
${(z + x)}^{2} - y^{2} = 36 \Rightarrow (z + x + y) (z + x - y) = 36$
$\Rightarrow (z + x - y) = \frac{36}{x + y + z} . . . . (i i i)$
Adding $(i), (i i), (i i i)$ we get
$(x + y - z) + (y + z - x) + (z + x - y) = \frac{4}{x + y + z} + \frac{9}{x + y + z} + \frac{36}{x + y + z}$
$\Rightarrow (x + y + z) = \frac{49}{(x + y + z)}$
$\Rightarrow {(x + y + z)}^{2} = 49$
$\Rightarrow$ $x + y + z = 7$

Suggest Corrections

Similar questions

x, y, z

x + log (x + \sqrt{x^{2} + 1}) = y

y + log (y + \sqrt{y^{2} + 1}) = z

z + log (z + \sqrt{z^{2} + 1}) = x

{(1 - x)}^{2} + {(1 - y)}^{2} + {(1 - z)}^{2}

\frac{1}{x + 1} + \frac{2}{y + z} + \frac{2006}{2006} = 1

\frac{x^{2}}{x^{2} + x} + \frac{y^{2}}{y^{2} + y} + \frac{z^{2}}{z^{2} + 2006 z}

x + y + z = 0

\frac{1}{x^{2} + y^{2} - z^{2}} + \frac{1}{y^{2} + z^{2} - x^{2}} + \frac{1}{z^{2} + x^{2} - y^{2}}

x^{2} + y^{2} + z^{2} = 5, x, y, z \in R,

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

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

Join BYJU'S Learning Program