Question

12. If x+y+z= 9, then find the value of
(3-x)3 + (3-y)2 + (3-2)} – 3(3-x)(3-y)(3-2).​

Answer 1

Answer:

Explanation:

Since a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab−bc−ac)

In the expression to be evaluated, a=2-x,b=2-y,c=2-z and a+b+c=6-(x+y+z)

Given x+y+z=6, a+b+c=0,

So, a3+b3+c3–3abc=(0)(a2+b2+c2–ab−bc−ac)=0

If it is known that the answer is constant, you can also rely on checking with 3–4 test cases satisfying given equation such as(2,2,2),(1,2,3),(0,2,4),(0,0,6) etc.

Answer 2

Answer:

refer to the attachment