Question

expand (3/2x_ 4/9y)³​

Answer 1

Step-by-step explanation:

The standard algebraic identities are:

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

a2 – b2 = (a + b)(a – b)

(x + a)(x + b) = x2 + (a + b) x + ab

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(a + b)3 = a3 + b3 + 3ab (a + b)

(a – b)3 = a3 – b3 – 3ab (a – b)

a3 + b3 + c3– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Answer 2

We have been asked to expand :

$\small{\sf{({\dfrac{3}{2}x - \dfrac{4}{9}y})^{3}}}$

If we carefully observe ; it is in the form of (a - b)³ formula . So , we will use the same identity to solve this problem .

Let's first take the identity with its expansion :

$\small{ \sf{({a - b})^{3}}} \: = \sf{a}^{3} - {b}^{3} - 3 {a}^{2}b + 3a{b}^{2}$

The above given expansion is just the simplified form of :

$\small{ \sf{({a - b})^{3}}} \: = \sf{a}^{3} - {b}^{3} - 3ab(a - b)$

We can use either of them anytime ; but here we will use the first form .

Next we need to consider the values :

$\small{\longrightarrow{\sf{a = \dfrac{3}{2}x}}}$

$\small{\longrightarrow{\sf{b = \dfrac{4}{9}y}}}$

Substituting the values :

$\small{\longrightarrow{\sf{({\dfrac{3}{2}x - \dfrac{4}{9}y})^{3} = ({\dfrac{3}{2}}x)^{3}- ({\dfrac{4}{9}}y)^{3} - 3({\dfrac{3}{2}}x)^{2} \times \dfrac{4}{9}y + 3 \times \dfrac{3}{2}x \times ({\dfrac{4}{9}}^{2})y }}}$

Simplifying it ; we get :

$\small{\longrightarrow{\sf{{\dfrac{27}{8}x}^{3} - {\dfrac{64}{729}y}^{3} - 3 \times \dfrac{9}{4}x \times \dfrac{4}{9}y + 3 \times \dfrac{3}{2}x \times \dfrac{16}{81}y}}}$

$\small{\longrightarrow{\sf{\dfrac{27}{8}x}^{3} - {\dfrac{64}{729}y}^{3} - 3 \times \cancel{\dfrac{9}{4}x \times \dfrac{4}{9}}y + \cancel{3} \times \dfrac{8}{\cancel{27}}{y}^{2}}}$

$\large{\boxed{\implies{\sf{\pink{{\dfrac{27}{8}x}^{3} - {\dfrac{64}{729}y}^{3} - 3 xy+ \dfrac{8}{9}{xy}^{2}}}}}}$