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expanded form help 

critical  Sep 15, 2018
 #1
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+1

Expand the following:
(2 x - (3 y^2)/4)^3

 

(2 x - (3 y^2)/4)^3 | = | sum_(k=0)^3 binomial(3, k) ((-3 y^2)/4)^(3 - k) (2 x)^k | 
 | = | binomial(3, 0) ((-3 y^2)/4)^3 (2 x)^0 + binomial(3, 1) ((-3 y^2)/4)^2 (2 x)^1 + binomial(3, 2) ((-3 y^2)/4)^1 (2 x)^2 + binomial(3, 3) ((-3 y^2)/4)^0 (2 x)^3 | : invisible comma 
-27/64 binomial(3, 0) y^6 + 9/8 binomial(3, 1) x y^4 - 3 binomial(3, 2) x^2 y^2 + 8 binomial(3, 3) x^3

 

The binomial coeffients comprise the 4^th row of Pascal's triangle:
(2 x)^3 - (33 (2 x)^2 y^2)/4 + 2×3 x ((-3 y^2)/4)^2 + ((-3 y^2)/4)^3

Multiply each exponent in 2 x by 2:
(2 x)^3 - (3×3×2^2 x^2 y^2)/4 + 2×3 x ((-3 y^2)/4)^2 + ((-3 y^2)/4)^3

(2 x)^3 - (34×3 x^2 y^2)/4 + 2×3 x ((-3 y^2)/4)^2 + ((-3 y^2)/4)^3

(4 x^2 (-3) y^2×3)/4 = 4/4×x^2 (-3) y^2×3 = x^2 (-3) y^2×3:
(2 x)^3 + -3×3 x^2 y^2 + 2×3 x ((-3 y^2)/4)^2 + ((-3 y^2)/4)^3

 

Multiply each exponent in (-3 y^2)/4 by 2:
(2 x)^3 - 3×3 x^2 y^2 + 2×3 x ((-3)/4)^2 y^(2×2) + ((-3 y^2)/4)^3

(2 x)^3 - 3×3 x^2 y^2 + 2 ((-3)/4)^2 3 x y^4 + ((-3 y^2)/4)^3

((-3)/4)^2 = (-3)^2/4^2:
(2 x)^3 - 3×3 x^2 y^2 + 2×(-3)^2/4^2 3 x y^4 + ((-3 y^2)/4)^3

(2 x)^3 - 3×3 x^2 y^2 + (2×9×3 x y^4)/4^2 + ((-3 y^2)/4)^3

(2 x)^3 - 3×3 x^2 y^2 + (2×9×3 x y^4)/16 + ((-3 y^2)/4)^3

(2 x)^3 - 3×3 x^2 y^2 + (9×3 x y^4)/8 + ((-3 y^2)/4)^3

(2 x)^3 - 3×3 x^2 y^2 + (27 x y^4)/8 + ((-3 y^2)/4)^3

 

Multiply each exponent in 2 x by 3:

2^3 x^3 - 3×3 x^2 y^2 + (27 x y^4)/8 + ((-3 y^2)/4
2×2^2 x^3 - 3×3 x^2 y^2 + (27 x y^4)/8 + ((-3 y^2)/4)^3
2×4 x^3 - 3×3 x^2 y^2 + (27 x y^4)/8 + ((-3 y^2)/4)^3

8 x^3 - 3×3 x^2 y^2 + (27 x y^4)/8 + ((-3 y^2)/4
8 x^3 + -9 x^2 y^2 + (27 x y^4)/8 + ((-3 y^2)/4)^3

 

Multiply each exponent in (-3 y^2)/4 by 3:
8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 + ((-3)/4)^3 y^(3×2)

8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 + ((-3)/4)^3 y^6

8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 + (-3)^3/4^3 y^6

8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 + (-3^3 y^6)/4^
8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 - (3^3 y^6)/(4×4^2)
8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 - (3^3 y^6)/(4×16)

8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 - (3^3 y^6)/64

8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 - (3×3^2 y^6)/64

8 x^3 - 9 x^2 y^2 + (27 x y^4)/8 - (3×9 y^6)/64


8x^3 - 9x^2y^2 + (27xy^4)/8 - (27y^6)/64- or the 3rd on your list.

Guest Sep 15, 2018
 #2
avatar+27133 
+1

In general:  \((a-b)^3=a^3-3a^2b+3ab^2-b^3\)  so we can also do it like this:

 

\((2x-\frac{3}{4}y^2)^3\rightarrow (2x)^3-3(2x)^2(\frac{3}{4}y^2)+3(2x)(\frac{3}{4}y^2)^2-(\frac{3}{4}y^2)^3\\ \rightarrow 8x^3-9x^2y^2+\frac{27}{8}xy^4-\frac{27}{64}y^6\)

 

where \(a=2x \text{ and } b=\frac{3}{4}y^2\)

.

Alan  Sep 16, 2018

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