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Expand the following:

(5 x^2 - a b)^3

Expand (5 x^2 - a b)^3 using the binomial expansion theorem.

(5 x^2 - a b)^3 = sum_(k=0)^3 binomial(3, k) (5 x^2)^(3 - k) (-a b)^k = binomial(3, 0) (5 x^2)^3 (-a b)^0 + binomial(3, 1) (5 x^2)^2 (-a b)^1 + binomial(3, 2) (5 x^2)^1 (-a b)^2 + binomial(3, 3) (5 x^2)^0 (-a b)^3:

125 binomial(3, 0) x^6 - 25 binomial(3, 1) a b x^4 + 5 binomial(3, 2) a^2 b^2 x^2 - binomial(3, 3) a^3 b^3

Evaluate the binomial coefficients by looking at Pascal's triangle

(-a b)^3 + 5×3 (-a b)^2 x^2 - 3 a b (5 x^2)^2 + (5 x^2)^3

(-a b)^3 + 5×3×(-1)^2 a^2 b^2 x^2 - 3 a b (5 x^2)^2 + (5 x^2

(-a b)^3 + 5×3 a^2 b^2 x^2 - 3 a b (5 x^2)^2 + (5 x^2)^3

(-a b)^3 + 5×3 a^2 b^2 x^2 - 3 a b×5^2 x^(2×2) + (5 x^2)^3

(-a b)^3 + 5×3 a^2 b^2 x^2 - 5^2×3 a b x^4 + (5 x^2)^3

(-a b)^3 + 5×3 a^2 b^2 x^2 - 25×3 a b x^4 + (5 x^2)^3

(-1)^3 a^3 b^3 + 5×3 a^2 b^2 x^2 - 25×3 a b x^4 +

-1 a^3 b^3 + 5×3 a^2 b^2 x^2 - 25×3 a b x^4 + (5 x^2)^3

-(a^3 b^3) + 15 a^2 b^2 x^2 - 25×3 a b x^4 + (5 x^2)^3

-(a^3 b^3) + 15 a^2 b^2 x^2 + -75 a b x^4 + (5 x^2)^3

-(a^3 b^3) + 15 a^2 b^2 x^2 - 75 a b x^4 + 5^3 x^(3×2)

-(a^3 b^3) + 15 a^2 b^2 x^2 - 75 a b x^4 + 5^3 x^6

-(a^3 b^3) + 15 a^2 b^2 x^2 - 75 a b x^4 + 5×5^2 x^6

-(a^3 b^3) + 15 a^2 b^2 x^2 - 75 a b x^4 + 5×25 x^6

-(a^3 b^3) + 15a^2b^2x^2 - 75abx^4 + 125x^6

(5x^2  - ab)^3     

Let  a  =  5x^2         and  b  =   (ab)  

So     (a  - b)^3   =   a^3  -  3a^2b   +  3ab^2 -  b^3

So we have

1 * (5x^2)^3  -   3(5x^2)^2*ab  +  3(5x^2)*(ab)^2 -  1(ab)^3  =

125x^6  -  75x^4ab  +  15x^2(ab)^2  - (ab)^3