Question

If $a+b+c=0$, then find the value of $(a+b-c{)}^3+(b+c-a{)}^3+(c+a-b{)}^3$.$\left[3Marks\right]$

Answer 1

$a+b+c=0$
Now, $(a+b-c{)}^3=(a+b+c–c-c{)}^3$
$=(0–2c{)}^3=-8c^3$$\left[0.5Mark\right]$
$(b+c-a{)}^3=(b+c+a–a-a{)}^3$
$=(0–2a{)}^3=-8a^3$$\left[0.5Mark\right]$
$(c+a-b{)}^3=(c+a+b–b-b{)}^3$
$=(0–2b{)}^3=-8b^3$$\left[0.5Mark\right]$

$\therefore (a+b-c{)}^3+(b+c-a{)}^3+(c+a-b{)}^3$
$=(-8c^3)+(-8a^3)+(-8b^3)$
$=-8(a^3+b^3+c^3)=-8\left(3abc\right)=-24abc$....... using :[$a^3+b^3+c^3-3abc$$\left[1Mark\right]$

$=(a+b+c)(a^2+b^2+c^2–ab–bc-ca)$]
[If $a+b+c=0$ then $a^3+b^3+c^3=3abc]$$\left[0.5Mark\right]$