Question

If $2x+3y+z=15$, then value of $(5-2x{)}^3+(5-3y{)}^3+(5-z{)}^3-3(5-2x\left)\right(5-3y\left)\right(5-z)$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $3$

Answer 1

The correct option is C $0$

Let $5-2x=a,5-3y=b$ and $5-z=c$, then $(5-2x)+(5-3y)+(5-z)=a+b+c$
$\Rightarrow 15-(2x+3y+z)=a+b+c$
$\Rightarrow 15-15=a+b+c$ ...[$\therefore 2x+3y+z=15$ (given)]
$\therefore a+b+c=0$
Using the result : if $a+b+c=0$,then
$a^3+b^3+c^3-3abc$ = 0 .....(1)
Put back the values of $a,b$ and $c$ in (1), we get
$(5-2x{)}^3+(5-3y{)}^3+(5-z{)}^3-3(5-2x\left)\right(5-3y\left)\right(5-z)=0$
Hence, the value of $(5-2x{)}^3+(5-3y{)}^3+(5-z{)}^3-3(5-2x\left)\right(5-3y\left)\right(5-z)$ is 0.