Question

If x+y+z=9,then find the value of (3-x)³+(3-y)³+(3-z)³-3(3-x)(3-y)(3-z)

Answer 1

x+y+z= 9

It can be written as

(3-x)+(3-y)+(3-z) = 0


By the above result


The value of the given expression is 0

Answer 2

Answer:

0

Step-by-step explanation:

It is given that

$x+y+z=9$

Let $a=(3-x),b=(3-y),c=(3-z)$. So, $x=(3-a),y=(3-b),z=(3-c)$.

$(3-a)+(3-b)+(3-c)=9$

$3-a+3-b+3-c=9$

$9-a-b-c=9$

Subtract 9 from both sides.

$-a-b-c=9-9$

$-(a+b+c)=0$

$a+b+c=0$

we need to find the value of

$(3-x)^3+(3-y)^3+(3-z)^3-3(3-x)(3-y)(3-z)$

Substitute $(3-x)=a,(3-y)=b,(3-z)=c$,

$a^3+b^3+c^3-3abc$

We know that if a+b+c=0, then $a^3+b^3+c^3-3abc=0$.

$a^3+b^3+c^3-3abc=0$

Therefore the value of given expression is 0.