Perfect Cube of a Binomial: Extended Formulas

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Cube of a sum of two terms:

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

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

Or: $(a + b)^{3} = (a + b) (a + b)^{2} = (a + b) (a^{2} + 2 ab + b^{2})$

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

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

(2 + 3)^{3} = 2^{3} + 3 \cdot 2^{2} \cdot 3 + 3 \cdot 2 \cdot 3^{2} + 3^{3}

= 8 + 36 + 54 + 27 = 125

And indeed: $5^{3} = 125$ ✓

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

= x^{3} + 6 x^{2} + 12 x + 8

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

(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1

Cube of a difference:

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

Signs alternate: $+ - + -$

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

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

(a + b)^{3} \neq = a^{3} + b^{3}

The middle terms $3 a^{2} b + 3 a b^{2}$ are missing!

(a - b)^{3} = a^{3} - b^{3} (WRONG!)

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

Practice cube of binomial: