Question

Check whether (x+3)whole cube=xcube-8 is a quadratic equation?

Answer 1

$(x + 3) ^{3} = x^{3} - 8 \\ using \: identity \: (a + b)^{3} \\ \: \: \\ = x ^{3} + 9x {}^{2} + 27x + 9 = x {}^{3} - 8 \\ = x {}^{3} - x {}^{3} + 9x {}^{2} + 27x + 9 + 8 \\ = 9x {}^{2} + 27x + 17 = 0 \\ hence \: (x + 3) {}^{3} = x {}^{3} - 8 \: is \: a \: quadratic \: equation$

Answer 2

Answer:

Yes, the given equation is a quadratic equation.

Step-by-step explanation:

The given equation is

$(x+3)^3=x^3-8$

We need to check whether the given equation is a quadratic equation or not.

Highest degree of a quadratic equation is 2.

Apply cubic formula.

$x^3+3(x)^2(3)+3(x)(3)^2+(3)^3=x^3-8$        $[\because(a+b)^3=a^3+3a^2b+3ab^2+b^3]$

$x^3+6x^2+27x+27=x^3-8$

Subtract x³ from both sides.

$x^3+6x^2+27x+27-x^3=x^3-8-x^3$

$6x^2+27x+27=-8$

Add 8 on both sides.

$6x^2+27x+27+8=-8+8$

$6x^2+27x+35=0$

The highest degree of this polynomial is 2.

Therefore, the given equation is a quadratic equation.