What is expand (x_{0}+αx^{1/2})^3 ?

The solution to expand (x_{0}+αx^{1/2})^3 is x_{0}^3+3x_{0}^2αx^{1/2}+3x_{0}α^2x+α^3x^{3/2}

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expand (x_{0}+αx^{1/2})^3

expand

(x_{0} + α x^{\frac{1}{2}})^{3}

x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} α^{2} x + α^{3} x^{\frac{3}{2}}

Solution steps

(x_{0} + α x^{\frac{1}{2}})^{3}

Apply Perfect Cube Formula:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

(x_{0} + α x^{\frac{1}{2}})^{3} = x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} (α x^{\frac{1}{2}})^{2} + (α x^{\frac{1}{2}})^{3}

= x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} (α x^{\frac{1}{2}})^{2} + (α x^{\frac{1}{2}})^{3}

(α x^{\frac{1}{2}})^{2} = α^{2} x

(α x^{\frac{1}{2}})^{3} = α^{3} x^{\frac{3}{2}}

= x_{0}^{3} + 3 x_{0}^{2} α x^{\frac{1}{2}} + 3 x_{0} α^{2} x + α^{3} x^{\frac{3}{2}}

\frac{x ^{3} + 1}{( x - 2 ) ^{2}}

\frac{a ( a - 1 )}{2}

a (2 \cdot 3 + a)

(2 - \frac{x}{2})^{7}

\frac{8}{x ( x - 2 )}

What is expand (x_{0}+αx^{1/2})^3 ?
The solution to expand (x_{0}+αx^{1/2})^3 is x_{0}^3+3x_{0}^2αx^{1/2}+3x_{0}α^2x+α^3x^{3/2}