What is expand (1-x^{1/2})^3 ?

The solution to expand (1-x^{1/2})^3 is 1-3x^{1/2}+3x-x^{3/2}

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expand (1-x^{1/2})^3

expand

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

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

Solution steps

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

Apply Perfect Cube Formula:

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

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

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

1^{3} = 1

3 \cdot 1^{2} x^{\frac{1}{2}} = 3 x^{\frac{1}{2}}

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

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

= 1 - 3 x^{\frac{1}{2}} + 3 x - x^{\frac{3}{2}}

(1 - tan^{2} (x)) (1 - tan^{2} (2 x)) tan (4 x)

(x^{\frac{1}{3}} - \frac{1}{4} x^{- \frac{1}{3}})^{2}

tan (h) (x - y)

(x + ln (x))^{2}

lo g_{9 (\frac{3 ^{3}}{7})^{4}} (x)

What is expand (1-x^{1/2})^3 ?
The solution to expand (1-x^{1/2})^3 is 1-3x^{1/2}+3x-x^{3/2}