SSC CHSL Model Paper 13

Question : 68 of 100

Marks: +1, -0

If a + b + c = 11 and ab + be + ca = 38, then

a^3 + b^3 + c^3 - 3abc

is equal to:

Solution:

a + b + c = 11

(a + b + c)^2 = (11)^2

a^2 + b^2 + c^2

+ 2(ab + bc + ca) =121

⇒ a^2 + b^2 + c^2 + 2 × 38 = 121

⇒ a^2 + b^2 + c^2 = 121 - 76

⇒ a^2 +b^2 + c^2 = 45

a^3+b^3+c^3-3abc

=(a+b+c)

[a^2+b^2+c^2

-(ab+bc+ca)]

=11[45-38]

=11×7=77

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