Question

Help Me Guys⚡️⚡️....
Please Do It Fast...✌️✌️

Answer 1

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Hi GOOD Morning HERE UR ANSWER!!!!!!!
Given equation is (x+3)^3=x^3-8

Expand (x+3)^3 using (a+b)^3=a^3+b^3+3a^2b+3ab^2

So, using this expand it

(x+3)^3=x^3+3^3+3×x^2×3+3×x×3^2

=x^3+27+9x^2+27x

AGAIN REPLACE in equation now

x^3+9x^2+27x+27=x^3-8

Here both side of eq has x^3 both so they got cancel.

9x^2+27x+27+8=0

9x^2+27x+35=0

AGAIN, we know that in quadratic equation the highest degree is 2

So, here the highest degree is also 2

so, it is a $\red{\huge{quadratic\: equation}}$

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