Question

please provide it's solution​

Answer 1

Step-by-step explanation:

$\frac{ {(a - b)}^{3} + 8 {a}^{3} }{3 {a}^{2} + {b}^{2} } = 3a - b \\ {(a - b)}^{3} + 8 {a}^{3} = (3a - b)(3 {a}^{2} + {b}^{2}) \\ {a}^{3} - {b}^{3} - 3 {a}^{2}b + 3 a{b}^{2} + 8 {a}^{3} =9 {a}^{3} + 3a {b}^{2} - 3 {a}^{2} b - {b}^{3} \\ 9 {a}^{3} - {b}^{3} - 3 {a}^{2} b + 3a {b}^{3} = 9 {a}^{3} - {b}^{3} - 3 {a}^{2} b + 3a {b}^{3} \\ L.H.S=R.H.S \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$

Answer 2

Step-by-step explanation:

Given:-

a,b,c are three non zero real numbers and

a+b+c = 0

To find :-

Prove that : [(a-b)³+8a³]/(3a²+b²) = 3a-b

Solution :-

Given that

a,b,c are non-zero real numbers

a+b+c = 0

We know that

(a+b)³ = a³+b³+3a²b+3ab²

(a-b)³ = a³-b³-3a²b+3ab²

On taking LHS :-

[(a-b)³+8a³]/(3a²+b²)

=> [ (a³-b³-3a²b+3ab²)+8a³]/(3a²+b²)

=> (a³-b³-3a²b+3ab²+8a³)/(3a²+b²)

=> (9a³-b³-3a²b+3ab²)/(3a²+b²)

It can be rearranged as

=> [(9a³+3ab²)-(b³+3a²b)]/(3a²+b²)

=> [3a(3a²+b²) -b(b²+3a²) ] /(3a²+b²)

=>[3a(3a²+b²) -b(3a²+b²) ] /(3a²+b²)

=> (3a²+b²)(3a-b)/(3a²+b²)

On cancelling (3a²+b²) in both the numerator and the denominator then

=> 3a-b

=> LHS = 3a-b

=> RHS

=> LHS = RHS

Answer :-

If a,b,c are three non zero real numbers and a+b+c = 0 then

[(a-b)³+8a³]/(3a²+b²) = 3a-b

Used formulae:-

• (a-b)³ = a³-b³-3a²b+3ab²