Delhi Police Constable Exam Model Paper 4

Tips & Tricks
GIVEN:
${\left(\mathit{x}\mathit{–}3\right)}^2+{\left(\mathit{y}+5\right)}^2+{\left(\mathit{z}\mathit{–}2\right)}^2=0$
FORMULA USED:
If x + y + z = 0, then ${\mathit{x}}^3+{\mathit{y}}^3+{\mathit{z}}^3=3\mathit{x}\mathit{y}\mathit{z}$
CALCULATION:
Sum of square will be zero ‘0’
only and when they are individually zero ‘0’.
$\mathit{x}\mathit{–}3=0$
⇒ x = 3And, y + 5 = 0
⇒ y = -5
And, z – 2 = 0
⇒ z = 2
Since
$\mathit{x}+\mathit{y}+\mathit{z}=3+\left(-5\right)+2=0$
Hence, ${\mathit{x}}^3+{\mathit{y}}^3+{\mathit{z}}^3=3\mathit{x}\mathit{y}\mathit{z}$
$=3\times 3\times \left(-5\right)\times 2=-90$