Question

a3 +b3+8c3-6abc factorise​

Answer 1

Answer:

Step-by-step explanation:

$a^{3}$ + $b^{3}$ + $8c^{3}$ - 6abc

= $(a)^{3}$ + $b^{3}$ + $(2c)^{3}$ - 3(a * b * 2c)

= ( a + b + c) {$a^{2}$ + $b^{2}$ + $(2c)^{2}$ - ab - 2bc - 2ca}

= (a + b+ c) ( $a^{2}$ + $b^{2}$ + $4c^{2}$ - ab - 2bc - 2ca)

Answer 2

a3 +b3 +8c3 - 6abc

=(a +b )3 - 3ab(a + b ) + c3 - 6abc(a + b + c) - 3abc (a + b) c (a+ b + c) - 3ab(a + b) - 6abc

= (a + b + c)3 - 3(a +b) c(a + b + c) - 3ab( a + b) - (3 +3)abc

=( a + b + c) 3 - 3c[ ( a+b )(a + b + c) ] - 3ab(a + b + c)

now factor (a + b + c) and go on

( a + b + c)[(a + b + c)2 - 3c(a + b) - 3ab) ]

= (a + b + c)(a2 +(6)2 + c2 + 2ab + 2ac + 2ac - 3ab - 3bc - 3ac)

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

I hope it helps you

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