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MAT Exam 2023 Paper

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Question : 118 of 200

Marks: +1, -0

If

(x+y+z)=12

\left(x^{2}+y^{2}+z^{2}\right)=50

, find the value of

\left(x^{3}+y^{3}+z^{3}-3 x y z\right)

36
24
48
42

Solution:

Concept:

Use the standard identities involving

(x+y+z)^2

and

x^3+y^3+z^3-3xyz

.

Explanation:

We are given

x+y+z=12

and

x^2+y^2+z^2=50

.

First, find

xy+yz+zx

using the square identity:

(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)

Substitute the given values:

(12)^2 = 50 + 2(xy+yz+zx)

144 = 50 + 2(xy+yz+zx)

xy+yz+zx = \frac{144-50}{2} = 47

Now apply the identity:

x^3+y^3+z^3-3xyz = (x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)

Substitute the known values:

=12\left(50-47\right) = 12 \times 3 = 36

Answer:

36

Correct option: A. 36

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