Question

If (a + 2b + c) = 0, then show that a^3 + 8b^3 + c^3 = 6abc.​

Answer 1

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$\mathfrak{The\:Answer\:is}$

(a + 2b + c) = 0

(a + 2b + c)³ = 0³ [Cubing both sides]

=>(a + 2b)³ + 3(a + 2b)²c + 3(a + 2b)c² + c³ = 0

=>(a³ + 6a²b + 12ab² + 8b³) + 3c(a + 2b)(a + 2b + c) + c³ = 0

=> (a³ + 8b³ + c³) + 6ab(a + 2b) = 0

=> a³ + 8b³ + c³ - 6abc = 0

=>a³ + 8b³ + c³ = 6abc. [Hence, proved]

$\boxed{Hope\:This\:Helps}$

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Answer 2

❏ Question:-

Q) if (a+2b+c)=0 then,

prove that , a³+8b³+c³=6abc.

❏ Solution:-

❚➾

Now , we know that ,

(X³+Y³+Z³)-3XYZ=(X+Y+Z)(X²+Y²+Z²-XY-YZ-ZX)

•Now if, (X+Y+Z)=0, then;

(X³+Y³+Z³)-3XYZ=0×(X²+Y²+Z²-XY-YZ-ZX)

➾(X³+Y³+Z³)-3XYZ=0

➾(X³+Y³+Z³)=3XYZ [$\bf\because X+Y+Z=0$ ]

❚ Now , from the Given problem,.

•(a+2b+c)=0

$\sf \therefore L.H.S.= a{}^{3}+8b{}^{3}+c{}^{3}$

$\sf = a{}^{3}+2{}^{3}.b{}^{3}+{c}^{3}$

$\sf =a{}^{3}+{2b}^{3}+{c}^{3}$

[ Now, by using the above property,

as (a+2b+c)=0, so ]

$\sf =3(a)(2b)(c)$

$\sf\boxed{\red{ =6abc=R.H.S.\:\:\:(proved)}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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