What is expand (L^{1/3}+K^{1/3})^3 ?

The solution to expand (L^{1/3}+K^{1/3})^3 is L+3L^{2/3}K^{1/3}+3L^{1/3}K^{2/3}+K

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expand (L^{1/3}+K^{1/3})^3

expand

(L^{\frac{1}{3}} + K^{\frac{1}{3}})^{3}

L + 3 L^{\frac{2}{3}} K^{\frac{1}{3}} + 3 L^{\frac{1}{3}} K^{\frac{2}{3}} + K

Solution steps

(L^{\frac{1}{3}} + K^{\frac{1}{3}})^{3}

Apply Perfect Cube Formula:

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

(L^{\frac{1}{3}} + K^{\frac{1}{3}})^{3} = (L^{\frac{1}{3}})^{3} + 3 (L^{\frac{1}{3}})^{2} K^{\frac{1}{3}} + 3 L^{\frac{1}{3}} (K^{\frac{1}{3}})^{2} + (K^{\frac{1}{3}})^{3}

= (L^{\frac{1}{3}})^{3} + 3 (L^{\frac{1}{3}})^{2} K^{\frac{1}{3}} + 3 L^{\frac{1}{3}} (K^{\frac{1}{3}})^{2} + (K^{\frac{1}{3}})^{3}

(L^{\frac{1}{3}})^{3} = L

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

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

(K^{\frac{1}{3}})^{3} = K

= L + 3 L^{\frac{2}{3}} K^{\frac{1}{3}} + 3 L^{\frac{1}{3}} K^{\frac{2}{3}} + K

(1 + cos (x))^{3}

(cos (x) + y sin (x))^{2}

P (2, - 3) A (c + 2, - 4) B (4 - 3 c, - 2)

l im x \Rightarrow 2 (3 x^{3} - 2 x^{2} + 6 x)

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