Bihar Police Constable Model Paper 5

Given:
a = 2
b = 3
c = 4
Concept used:
${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3$ – 3abc = (1/2)(a + b + c) $\left({\left(\mathit{a}\mathit{–}\mathit{b}\right)}^2+{\left(\mathit{b}\mathit{–}\mathit{c}\right)}^2+{\left(\mathit{c}\mathit{–}\mathit{a}\right)}^2\right)$
Calculations:
${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3$ – 3abc = (1/2)(2 + 3 + 4) $\left({\left(2\mathit{–}3\right)}^2+{\left(3\mathit{–}4\right)}^2+{\left(4\mathit{–}2\right)}^2\right)$
⇒ (9/2)(1 + 1 + 4)
⇒ 9 × 3 = 27
∴ The value of ${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3$ – 3abc is 27