Question

• If a+b+2c=0 then prove a³+b³+8c³ = 6abc​

Answer 1

Answer:

proof below

Step-by-step explanation:

a+b=-2c

(a+b)³ = (-2c)³

a³+b³+3ab(a+b)=-8c³

a³+b³+8c³=-3ab(-2c) {as a+b=-2c}

a³+b³+8c³=6abc

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Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:a + b + 2c = 0$

$\large\underline{\sf{To\:prove - }}$

$\red{\rm :\longmapsto\: {a}^{3} + {b}^{3} + {8c}^{3} = 6abc}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a + b + 2c = 0$

We know that,

$\red{\rm :\longmapsto\:If \: a + b + c = 0 \: then \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc}$

So, using this identity, we get

${\rm :\longmapsto\:{a}^{3} + {b}^{3} + {(2c)}^{3} = 3ab(2c)}$

${\rm :\longmapsto\:{a}^{3} + {b}^{3} + {8c}^{3} = 6abc}$

Hence, Proved

Alternative Method :-

Given that,

$\rm :\longmapsto\:a + b + 2c = 0$

can be rewritten as

$\rm :\longmapsto\:a + b = - 2c$

On Cubing both sides, we get

$\rm :\longmapsto\:(a + b) {}^{3} = ( - 2c) {}^{3}$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + 3ab(a + b) = - 8 {c}^{3}$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + 3ab( - 2c) = - 8 {c}^{3} \: \: \: \{ \because \: a + b = - 2c \}$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} - 6abc = - 8 {c}^{3} \: \: \: \{ \because \: a + b = - 2c \}$

$\rm :\longmapsto\: {a}^{3} + {b}^{3} + 8 {c}^{3} = 6abc$

Hence, Proved

Additional Information :-

More Identities to know :-

$\boxed{ \rm{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \rm{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\boxed{ \rm{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$\boxed{ \rm{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\boxed{ \rm{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}}}$

$\boxed{ \rm{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}}}$