Question

Simplify the expression

( 3^(1÷5) + 3^(1÷2) )÷( (1-3^(3÷5) * 3^(1÷5) )
Note: the answer is
1 ÷ (1 - 3^(3÷10) )
Comment: It's mind-boggling how the author of the exercise's book used the term "simplify" then shipped the answer with the radicand in denominator I just can't find a common term to get this expression in the answer form

Answer 1

`\large\text{$\frac{1}{1 - 3^{(3)/(10)}}$}`

Explanation:

Given rational expression:

`\large\text{$(3^(\frac15) + 3^(\frac12))/((1-3^(\frac35))\cdot 3^(\frac15))$}`

Rewrite the exponent 1/2 in the numerator as the sum 1/5 + 3/10:

`\large\text{$\frac{3^(\frac15) + 3^{\frac15+(3)/(10)}}{(1-3^(\frac35))\cdot 3^(\frac15)}$}`

`\textsf{Apply the exponent rule:} \quad a^(b+c)=a^b \cdot a^c`

`\large\text{$(3^(\frac15) + 3^(\frac15)\cdot 3^(3)/(10))/((1-3^(\frac35))\cdot 3^(\frac15))$}`

Factor out
`3^(\frac15)` from the numerator:

`\large\text{$\frac{3^(\frac15) (1+ 3^{(3)/(10)})}{(1-3^(\frac35))\cdot 3^(\frac15)}$}`

Cancel the common factor
`3^(\frac15)`:

`\large\text{$\frac{1+ 3^{(3)/(10)}}{1-3^(\frac35)}$}`

Rewrite the exponent 3/5 in the denominator as the product of 3/10 and 2:

`\large\text{$\frac{1+ 3^{(3)/(10)}}{1-3^{((3)/(10)\cdot 2)}}$}`

`\textsf{Apply the exponent rule:} \quad a^(b\cdot c)=(a^b)^c`

`\large\text{$\frac{1+ 3^{(3)/(10)}}{1-\left(3^{(3)/(10)}\right)^2}$}`

We can now factor the denominator by applying the difference of two squares: a² - b² = (a + b)(a - b).

In this case:

• 
  `a = 1`

• 
  `b = 3^{(3)/(10)}`

Therefore:

`\large\text{$\frac{1+ 3^{(3)/(10)}}{(1+ 3^{(3)/(10)})(1 - 3^{(3)/(10)})}$}`

Cancel the common factor
`(1+ 3^{(3)/(10)})`:

`\large\text{$\frac{1}{1 - 3^{(3)/(10)}}$}`