Question

plzz explain this question ​

Answer 1

Answer:

Use the formula

(a-b)³=a³-b³-3a²b+3ab²

Answer 2

Proving:

Step-by-step explanation:

$\ Given \ value: \\\\\ p \ =2 -a \\\\\ Prove: \\\\a^3+6ap+p^3-8=0 \\\\\ Solution: \\\\\ put \ the \ value \ of \ p \ in \ the \ given \ equation: \\\\a^3+6a(2-a)+(2-a)^3-8=0 \\\\ \ solve \ L.H.S \ part: \\\\ \ formula: \\\\ \therefore (a-b)^3 \ = \ a^3\ -\ b^3\ -\ 3a^2b +3ab^2 \\\\\rightarrow a^3+6a(2-a)+(2-a)^3-8=0 \\\\\rightarrow a^3+12a-6a^2+(2)^3-(a)^3-3.2^2.a+3.2.a^2 -8=0 \\\\\rightarrow a^3+12a-6a^2+8-a^3-12a+6a^2-8=0 \\\\\rightarrow 8-8=0 \\\\\rightarrow 0=0 \\\\L.H.S = R.H.S$

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