Question

If P = 2 - a , prove that a² + 6ap + p³ - 8 =0​

Answer 1

$\huge{\color{teal}{\underbrace{\textsf{\textbf{\color{lime}{Question:-}}}}}}$

If P = 2-a, prove that a³+6aP+P³-8=0.

$\huge{\color{teal}{\underbrace{\textsf{\textbf{\color{lime}{Solution:-}}}}}}$

$\rm a³ + 6aP + P³ - 8 = 0 ; P = 2 - a$

$\rm a³+6aP+P³-8= RHS$

$\rm 0 = LHS$

We have to prove that $\rm a³ + 6aP + P³ - 8 = 0$, so substitute $\rm P = 2 - a$ in it :

$\rm \mapsto a³ + 6(2-a)P + (2-a)³-8=0$

$\rm \mapsto (2-a)³=(2)³-(a)³-3(2)(a)(2-a)-8 \\ \rm \mapsto We \: used \: the \: identity \: (x-y)³=x³-y³-3(x)(y)(x-y)$

$\rm \mapsto a³+12a-6a²+8-a³-6a(2-a)-8=0$

$\rm \mapsto a³+12a-6a²+8-a³-12a+6a²-8=0$

$\rm \mapsto \cancel{a³}\cancel{+a³}\cancel{-6a²}\cancel{+6a²}\cancel{+12a}\cancel{-12a}\cancel{+8}\cancel{-8}=0$

$\rm \mapsto 0=0$

$\rm \mapsto LHS = RHS$

♨️♨️♨️♨️♨️♨️♨️♨️♨️

Hope it helps you :)

Useful ? Then mark it as Brainliest ✌️