Question

factorise : 1+b^3+8c^3-6bc ​

Answer 1

Answer:

We know the identity a

3

+b

3

+c

3

−3abc=(a+b+c)(a

2

+b

2

+c

2

−ab−bc−ca)

Using the above identity taking a=1,b=b and c=2c, the equation 1+b

3

+8c

3

−6bc can be factorised as follows:

1+b

3

+8c

3

−6bc=(1)

3

+(b)

3

+(2c)

3

−3(1)(b)(2c)

=(1+b+2c)[(1)

2

+(b)

2

+(2c)

2

−(1×b)−(b×2c)−(2c×1)]=(1+b+2c)(1+b

2

+4c

2

−b−2bc−2c)

Hence, a+b

3

+8c

3

−6bc=(1+b+2c)(1+b

2

+4c

2

−b−2bc−2c)

Answer 2

$\large\purple{\underline{{\boxed{\textbf{your answer}}}}}$

$1 + b {}^{3} + 8c {}^{3} - 6bc \\ it \: would \: remain \: same \: \\ it \: can \: be \: factorise \: \\ it \: is \: in \: its \: on \: form$

$\large\orange{\underline{{\boxed{\textbf{drop \: some \: thanks}}}}}$

I too would thanks your answer.