SSC CHSL 15 Nov 2015 Shift 1 Paper

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Question : 150 of 199

Marks: +1, -0

If

(x + y + z) =6

and

(xy + yz + zx)=10

, then the value of

x ^3 + y ^3 + z ^3 − 3xyz

is:

48
40
42
36

Solution:

Given :

(x + y + z) =6

and

(xy + yz + zx)=10

------(i)
Now,

(x + y + z) ^2 = (6) ^2

⇒

x^2 + y^2 + z^ 2

+ 2(xy + yz + zx) = 36

⇒

x ^2 + y ^2 + z^ 2 + 2(10) = 36

⇒

x ^2 + y ^2 + z^ 2 + = 36-20=16

-----(ii)
To find :

x^3+ y ^3 + z^ 3 − 3xyz

=

(x + y + z)(x ^2 + y ^2

+ z^ 2 − xy − yz − zx)

=

(6)(16 − 10)

=

6 × 6 = 36

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