Find $x^{2} + y^{2} + z^{2}$ if $x^{2} + x y + x z = 135, y^{2} + y z + y x = 351$ and $z^{2} + z x + z y = 243$ .

225

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

250

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

275

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

300

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

Open in App

Solution

The correct option is B $275$
Given : $x^{2} + x y + x z = 135$ --- ( 1 ) ,

$y^{2} + y z + y x = 351$ --- ( 2 )

And

$z^{2} + z x + z y = 243$ --- ( 3 )

Now we add equation 1 , 2 and 3 and get

$x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x = 729$

$(x + y + z)^{2} = 729$ ( We know : $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x)$

$(x + y + z)^{2} = 272$

$x + y + z = 27$ --- ( 4 )

Now we take equation 1 : $x^{2} + x y + x z = 135$

$x (x + y + z) = 135$ , Substitute value from equation 4 and get

$27 x = 135$

$x = 5$

And we take equation 2 : $y^{2} + y z + y x = 351$

$y (x + y + z) = 351$ , Substitute value from equation 4 and get

$27 y = 351$

$y = 13$

And we take equation 3 : $z^{2} + z x + z y = 243$

$z (x + y + z) = 243$ , Substitute value from equation 4 and get

$27 z = 243$

$z = 9$

Therefore,

Value of $x^{2} + y^{2} + z^{2} = 52 + 132 + 92 = 25 + 169 + 81 = 275$

Suggest Corrections

Similar questions

x^{2} + x y + x z = 135, y^{2} + y z + x y = 351

z^{2} + x z + y z = 243

x^{2} + y^{2} + z^{2} = ?

x^{2} + x y + x z = 135, y^{2} + y z + x y = 351

z^{2} + z x + z y = 243

(x, y, z > 0)

x^{2} + x y + x z = 18,

y^{2} + y z + y x + 12 = 0,

z^{2} + z x + z y = 30.

x^{2} + x y + x z = 18

y^{2} + y z + y z + 12 = 0

z^{2} + z x + z y = 30

Join BYJU'S Learning Program