Question

If $a+b+c=5$ and $ab+bc+ca=10$, then $a^3+b^3+c^3-3abc$ is equal to

• A. $-25$
• B. $25$
• C. $-20$
• D. $-15$

Answer 1

The correct option is A $-25$

$(a+b+c{)}^3=[(a+b)+c{]}^3=(a+b{)}^3+3(a+b{)}^{2}c+3(a+b)c^2+c^3$
$=>(a+b+c{)}^3=(a^3+3a^{2}b+3a{b}^2+b^3)+3(a^2+2ab+b^2)c+3(a+b)c^2+c^3$
$=>(a+b+c{)}^3=a^3+b^3+c^3+3a^{2}b+3a^{2}c+3a{b}^2+3b^{2}c+3a{c}^2+3b{c}^2+6abc$
$=>(a+b+c{)}^3=(a^3+b^3+c^3)+(3a^{2}b+3a^{2}c+3abc)+(3a{b}^2+3b^{2}c+3abc)+(3a{c}^2+3b{c}^2+3abc)-3abc$
$=>(a+b+c{)}^3=(a^3+b^3+c^3)+3a(ab+ac+bc)+3b(ab+bc+ac)+3c(ac+bc+ab)-3abc$
$=>(a+b+c{)}^3=(a^3+b^3+c^3)+3(a+b+c\left)\right(ab+ac+bc)-3abc$
$=>(a+b+c{)}^3=(a^3+b^3+c^3)+3[(a+b+c)(ab+ac+bc)-3abc]$
Putting the value given in the question we get,
$(5{)}^3=(a^3+b^3+c^3)+3(5\left)\right(10)-3abc$
$=>(a^3+b^3+c^3)-3abc=(5{)}^3-3(5\left)\right(10)$
$=>(a^3+b^3+c^3)-3abc=125-150$
$=>(a^3+b^3+c^3)-3abc=-25$