Question

Write the factors of a + b + c3-abc​

Answer 1

Step-by-step explanation:

ANSWER

We know the identity a 3+b 3+c 3−3abc=(a+b+c)2+b 2+c 2--ab−bc−ca)

Using the above identity taking a=a,b=−b and c=−c, the equation a

3

−b

3

−c

3

−3abc can be factorised as follows:

a

3

−b

3

−c

3

−3abc=(a

3

)+(−b)

3

+(−c)

3

−3(a)(−b)(c)

=[a+(−b)+(−c)][a

2

+(−b)

2

+(−c)

2

−(a×−b)−(−b×−c)−(−c×a)]

=(a−b−c)(a

2

+b

2

+c

2

+ab−bc+ca)

Hence, a

3

−b

3

−c

3

−3abc=(a−b−c)(a

2

+b

2

+c

2

+ab−bc+ca)

Answer 2

$\large\bold\purple{Answer}$

Given =a3 + b 3+ c3-3abc

$\mapsto$Let 'x' be the factor

f(x)=0 (after taking x as a factor )

x=0

The factors of this will be

$\mapsto$a=b=c=1,2,3,4,−−−−n