Question

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Answer 1

Answer:

$\dfrac{4}{9}x^{5}y-\dfrac{4}{3}x^{4}y^{2}+\dfrac{4}{9}x^{3}y^{3}$

Step-by-step explanation:

Given expression:

$\dfrac{4}{9}x^2yx(x^2-3xy+y^2)$

Distribute:

$\implies \dfrac{4}{9}x^2yxx^2-\dfrac{4}{9}x^2yx3xy+\dfrac{4}{9}x^2yxy^2$

Re-order each term to collect like variables:

$\implies \dfrac{4}{9}x^2x^2xy-\dfrac{4 \cdot 3}{9}x^2xxyy+\dfrac{4}{9}x^2xy^2y$

$\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^{b+c}:$

(Remembering that a = a¹)

$\implies \dfrac{4}{9}x^{2+2+1}y-\dfrac{12}{9}x^{2+1+1}y^{1+1}+\dfrac{4}{9}x^{2+1}y^{2+1}$

$\implies \dfrac{4}{9}x^{5}y-\dfrac{12}{9}x^{4}y^{2}+\dfrac{4}{9}x^{3}y^{3}$

Reduce 12/9 to 4/3:

$\implies \dfrac{4}{9}x^{5}y-\dfrac{4}{3}x^{4}y^{2}+\dfrac{4}{9}x^{3}y^{3}$

Learn more about exponent laws here:

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Answer 2

Answer: Hi I will solve this right quick.

Step-by-step explanation:

The equation is $\frac{4}{9} x^{2} y x(x^{2} -3xy+y^{2} )$

The answer is this

$\frac{4y (x^{2} - 3xy +y2) x^{3} }{9}$