What is expand (x^2+Δx^2-2)^3 ?

The solution to expand (x^2+Δx^2-2)^3 is Δ^3x^6+3Δ^2x^6+3Δx^6+x^6-6Δ^2x^4-12Δx^4-6x^4+12Δx^2+12x^2-8

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expand (x^2+Δx^2-2)^3

expand

(x^{2} + Δ x^{2} - 2)^{3}

Δ^{3} x^{6} + 3 Δ^{2} x^{6} + 3Δ x^{6} + x^{6} - 6 Δ^{2} x^{4} - 12Δ x^{4} - 6 x^{4} + 12Δ x^{2} + 12 x^{2} - 8

Solution steps

(x^{2} + Δ x^{2} - 2)^{3}

(x^{2} + Δ x^{2} - 2)^{3} = (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2)

= (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2)

Expand

(x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) (x^{2} + Δ x^{2} - 2) : Δ^{3} x^{6} + 3 Δ^{2} x^{6} + 3Δ x^{6} + x^{6} - 6 Δ^{2} x^{4} - 12Δ x^{4} - 6 x^{4} + 12Δ x^{2} + 12 x^{2} - 8

= Δ^{3} x^{6} + 3 Δ^{2} x^{6} + 3Δ x^{6} + x^{6} - 6 Δ^{2} x^{4} - 12Δ x^{4} - 6 x^{4} + 12Δ x^{2} + 12 x^{2} - 8

- 4 (x + Δ x)^{2}

(1 + e^{4 x})^{2}

- \frac{3 x}{h ^{2} ( x + h )}

T (3, - 4, 5)

- \frac{( t ^{2} + 1 ) ^{2}}{4 t}

What is expand (x^2+Δx^2-2)^3 ?
The solution to expand (x^2+Δx^2-2)^3 is Δ^3x^6+3Δ^2x^6+3Δx^6+x^6-6Δ^2x^4-12Δx^4-6x^4+12Δx^2+12x^2-8