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Parul Verma

· started a discussion

· 1 Months ago

1st (Question:108030.0)

value putting se kese

Question:

If x + y + z = 6, then the value of (x–1)3 + (y–2)3 + (z–3)3 is:

Options:
A) 2 (x–1) (y–2) (z–3)
B) 3(x–1) (y–2) (z–3)
C) 3xyz
D) (x–1) (y–2) (z–3)
Solution:
Ans: (b)

x + y + z = 6 (Given)

(x – 1) + (y – 2) + (z – 3) = x + y + z – 6 = 0

\(\therefore\) (x – 1)3 + (y – 2)3 + (z – 3)3 = 3(x – 1) (y – 2) (z – 3)

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