Question

Can somebody please help me simplify for number 8SHOULD BE POSITIVE EXPONENT

Answer 1

The expression is given as,

`\lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4`

Distributing the power inside the bracket ,

`\begin{gathered} \lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4\text{ = }\lbrack(x^3y^(-4))\text{ }*(2^(-1)x^(-3)y^(-3))^{}\rbrack^4 \\ \lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4\text{ = }\lbrack(x^3y^(-4))^4\text{ }*(2^(-1)x^(-3)y^(-3))^4\rbrack \end{gathered}`

Simplifying further,

`\lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4\text{ = }\lbrack(x^(12)y^(-16))*(2^(-4)x^(-12)y^(-12))\rbrack`

Rearranging the like terms,

`\lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4=16^(-1)*(x^(12)\text{ }*\text{ }x^(-12))*\text{ (}y^(-16)\text{ }*\text{ }y^(-12))`

By using law of exponents,

`\begin{gathered} x^a* x^b=x^(a+b) \\ x^0=\text{ 1} \end{gathered}`

Therefore,

`\begin{gathered} \lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4=16^(-1)\text{ }* x^(12-12)* y^(-16-12) \\ \lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4=16^(-1)\text{ }* x^0* y^(-28) \end{gathered}`

Further,

`\begin{gathered} \lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4=16^(-1)\text{ }y^(-28) \\ \lbrack(x^3y^(-4))\text{ }*(2x^3y^3)^(-1)\rbrack^4=\text{ }(1)/(16y^(28)) \end{gathered}`