Question

Factorize a^3+b^3+c^3-3abc by factor theorem? (a+b+c=0)

Answer 1

a³ + b³ + c³ -3abc

=a³ + (b³ + c³) -3abc

= a³ + (b +c)³ -3bc(b+c) -3abc

[ given, a+ b + c =0 => b+c = -a put it ]

= a³ + (b+c)³ -3bc(-a) -3abc

= a³ + (b + c)³ +3abc -3abc

[ use, formula , x³ + y³ = (x + y)(x² + y² -xy )

= {a + (b + c)}{a² + (b+c)² -a(b + c)}

=(a + b + c)(a² + b² + c² + 2bc - ab - ac)

Answer 2

Answer:

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+2bc-ab+ac)$

Step-by-step explanation:

Given : $a+b+c=0$

To find : Factorize $a^3+b^3+c^3-3abc$ by factor theorem?

Solution :

Solve the expression using factor theorem,

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

$=a^3+(b^3+c^3)-3abc$

Substitute, $-a=(b+c)$

$=a^3+(b^3+c^3)+3(b+c)bc$

Using formula, $(a+b)^3=a^3+b^3+3ab(a+b)$

$=a^3+(b+c)^3$

Using formula, $x^3+ y^3= (x + y)(x^2+ y^2-xy )$

$=(a+(b+c))[a^2+(b+c)^2-a(b+c)]$

$=(a+b+c)(a^2+b^2+c^2+2bc-ab+ac)$

Therefore, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+2bc-ab+ac)$