Question

refer to the attachment
No irrelevant answers please
thank you:)
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Answer 1

Answer: i.) $\frac{2x^{3}y }{6xy^{2} }$ = $\frac{2}{6}$ * $\frac{x^{3} }{x}$ * $\frac{y}{y^{2} }$

dividing the term

$\frac{2}{6}$ = $\frac{1}{3}$

$\frac{x^{3} }{x}$ = $x^{2}$

$\frac{y}{y^{2} }$ = $\frac{1}{y}$

$\frac{2x^{3}y }{6xy^{2} }$ = $\frac{x^{2} }{3y}$

ii.) $\sqrt[3]{216}$  =$\sqrt[3]{6^{3}}$

cancelling root 3 and power 3

$\sqrt[3]{216}$ = 6

iii.) $16^{\frac{3}{2} }$ - $16^{0}$

As $x^{0}$ = 1

$16^{\frac{3}{2} }$ - 1 =

$16^{\frac{3}{2} }$ = $4^{2\frac{3}{2} }$

multiplying the exponents

$4^{3}$ = 64

$16^{\frac{3}{2} }$ = 64

(iv.) $(\frac{3a^{2}b }{12ab^{4} }) ^{-2}$ = $(\frac{12ab^{4} }{3a^{2}b }) ^{2}$

solving the bracket

$(\frac{4b^{3} }{a})^{2}$ = $\frac{16b^{6} }{a^{2} }$

$(\frac{3a^{2}b }{12ab^{4} }) ^{-2}$ =  $\frac{16b^{6} }{a^{2} }$