If $p = 2 - a$ then prove that $a^{3} + 6 a p + p^{3} - 8 = 0$

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Solution

Given $p = 2 - a$

to prove $a^{3} + 6 a p + p^{3} - 8 = 0... (1)$

Now put the value of P in eq (1)

$a^{3} + 6 a (2 - a) + (2 - a)^{3} - 8 = 0$

$\Rightarrow a^{3} + 12 a - 6 a^{2} + 8 - a^{3} - 12 a + 6 a^{2} - 8 = 0$

$[∴ (a - b)^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}]$

$\Rightarrow 0 = 0$ (By eliminating equal terms we
have LHS = RHS)

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Similar questions

p = (2 - a)

a^{3} + 6 a p + p^{3} - 8

Find the value of a³+6ap+p³-8 when p=2-a

a + \frac{1}{a} = p

a \neq 0

a^{3} + \frac{1}{a^{3}} = p (p^{2} - 3)

p^{2} = a^{2} c o s^{2} θ + b^{2} s i n^{2} θ

P + \frac{d^{2} p}{d θ^{2}} = \frac{a^{2} b^{2}}{p^{3}}

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

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