Write the following cube in expanded form:
$(4x - 3y)^{3}$
$(4x - 3y)^{3}$
(N/A) To expand $(4x - 3y)^{3}$,we use the algebraic identity:
$(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b) = a^{3} - 3a^{2}b + 3ab^{2} - b^{3}$
Here,$a = 4x$ and $b = 3y$.
Substituting these values into the identity:
$(4x - 3y)^{3} = (4x)^{3} - 3(4x)^{2}(3y) + 3(4x)(3y)^{2} - (3y)^{3}$
$= 64x^{3} - 3(16x^{2})(3y) + 3(4x)(9y^{2}) - 27y^{3}$
$= 64x^{3} - 144x^{2}y + 108xy^{2} - 27y^{3}$
$(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b) = a^{3} - 3a^{2}b + 3ab^{2} - b^{3}$
Here,$a = 4x$ and $b = 3y$.
Substituting these values into the identity:
$(4x - 3y)^{3} = (4x)^{3} - 3(4x)^{2}(3y) + 3(4x)(3y)^{2} - (3y)^{3}$
$= 64x^{3} - 3(16x^{2})(3y) + 3(4x)(9y^{2}) - 27y^{3}$
$= 64x^{3} - 144x^{2}y + 108xy^{2} - 27y^{3}$
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