Question 88

If a+b+c=27, then what is the value of $$(a-7)^{3}+(b - 9)^{3}+(c - 11)^{3}-3(a - 7)(b - 9)(c - 11)$$ ?

Solution

If x+y+z = 0, $$(x)^{3}+(y)^{3}+(z)^{3}-3(x)(y)(z)=0$$

puty x = a-7, y = b-9, z = c-11,

then x+y+z = (a+b+c)-(7+11+9) = 27 - 27 = 0,

so $$(x)^{3}+(y)^{3}+(z)^{3}-3(x)(y)(z)=0$$

$$(a-7)^{3}+(b - 9)^{3}+(c - 11)^{3}-3(a - 7)(b - 9)(c - 11)=0$$

so the answer is option A.

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: