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Explanation:

Given, x + y + z = 10
Squaring both sides, we get
⇒ x2 + y2 + z2 + 2xy + 2yz + 2zx = 100
⇒ 30 + 2xy + 2yz + 2zx = 100
⇒ 2xy + 2yz + 2zx = 70
⇒ xy + yz + zx = 35

We know, x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

⇒ x3 + y3 + z3 - 3xyz = (10)(30 - 35)

⇒ x3 + y3 + z3 - 3xyz = (10)(-5)

⇒ x3 + y3 + z3 - 3xyz = -50

Hence, option (c).

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