Question

Factorise:a^3 - b^3 - C^3 - 3abc.​

Answer 1

We know that

a3 + b3+ c3− 3abc = (a + b + c)(a2 + b2+ c2− ab − bc − ca)………………………(1)

By using above equation let us consider

a = a, b= -b and c = -c

Then the equation (10 BECOMES

a3 + (-b3) + (-c3) − 3a(-b)(-c) = (a -b -c)(a2 + (-b2) + (-c2) − a(-b) − (-b)(-c) − (-c)a)

a3 -b3 -c3− 3abc = (a -b -c)(a2 +b2 +c2 + ab – bc + ca)

Hence a3 -b3 -c3 − 3abc = (a -b -c)(a2 +b2 +c2 + ab – bc + ca)

Answer 2

Answer:

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+2bc-ab+ac)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=0a+b+c=0

To find : Factorize a^3+b^3+c^3-3abca

3

+b

3

+c

3

−3abc by factor theore

Solve the expression using factor theorem,

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

Substitute, -a=(b+c)−a=(b+c)

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

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

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+b+c)(a^2+b^2+c^2+2bc-ab+ac)[/tex]

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