Question

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

Then prove that [(a-b)³+8a³]/(3a²+b²) = 3a-b​

Answer 1

To Prove : -

$\green{\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{[(a-b)³+8a³]}{(3a²+b²) }=3a-b}}}$

Proof :-

$\qquad$ ☀️We are given a,b,c are non-zero real numbers and a+b+c = 0.

Formula used :-

$\pink{\qquad\bigstar \quad \pmb {\mathfrak{(a+b)³ = a³+b³+3a²b+3ab²}}}$

$\pink{\qquad\bigstar\quad \pmb {\mathfrak{(a-b)³ = a³-b³-3a²b+3ab²}}}$

⠀⠀⠀⠀¯‎¯‎‎‎¯‎¯¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎‎¯‎‎¯‎‎‎¯‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎¯‎‎¯‎¯¯‎¯‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎

$\qquad\leadsto\quad \pmb {\mathfrak{L.H.S}}$

$\green{\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{[(a-b)³+8a³]}{(3a²+b²)}}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{ [ (a³-b³-3a²b+3ab²)+8a³]}{ (3a²+b²)}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{(a³-b³-3a²b+3ab²+8a³)}{(3a²+b²)}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{ (9a³-b³-3a²b+3ab²)}{(3a²+b²)}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ \dfrac{[(9a³+3ab²)-(b³+3a²b)]}{(3a²+b²)}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{[3a(3a²+b²) -b(b²+3a²) ]}{ (3a²+b²)}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{[3a(3a²+b²) -b(3a²+b²) ] }{(3a²+b²)}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{ (3a²+b²)(3a-b)}{(3a²+b²)}}}$

$\green{\qquad\leadsto\quad \pmb {\mathfrak{ 3a-b}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{R.H.S}}$

$\qquad$ $\qquad$ Proved..!! ☀️

⠀⠀⠀⠀¯‎¯‎‎‎¯‎¯¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎‎¯‎‎¯‎‎‎¯‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎¯‎‎¯‎¯¯‎¯‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎

Answer 2

Explanation:

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

a+b+c = 0

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