Where's One?

Just for sake of variety and completeness .....

Using the identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$ we have that

$\dfrac{8^{p} + 27^{p}}{12^{p} + 18^{p}} = \dfrac{2^{3p} + 3^{3p}}{2^{2p}3^{p} + 2^{p}3^{2p}} = \dfrac{(2^{p})^{3} + (3^{p})^{3}}{2^{p}3^{p}(2^{p} + 3^{p})} =$

$\dfrac{(2^{p} + 3^{p})(2^{2p} - 2^{p}3^{p} + 3^{2p})}{2^{p}3^{p}(2^{p} + 3^{p})} = \dfrac{2^{2p} - 2^{p}3^{p} + 3^{2p}}{2^{p}3^{p}} =$

$\left(\dfrac{2}{3}\right)^{p} - 1 + \left(\dfrac{3}{2}\right)^{p} = \dfrac{8 + 27}{12 + 18} = \dfrac{7}{6} \Longrightarrow \left(\dfrac{2}{3}\right)^{p} + \left(\dfrac{3}{2}\right)^{p} = \dfrac{13}{6}.$

Now letting $x = \left(\dfrac{2}{3}\right)^{p}$ our equation becomes

$x + \dfrac{1}{x} = \dfrac{13}{6} \Longrightarrow 6x^{2} - 13x + 6 = 0.$

By the quadratic equation we have that

$x = \dfrac{13 \pm \sqrt{13^{2} - 4*6*6}}{2*6} = \dfrac{13 \pm 5}{12},$

and so $x = \left(\dfrac{2}{3}\right)^{p}$ can equal either $\dfrac{13 - 5}{12} = \dfrac{2}{3}$ or $\dfrac{13 + 5}{12} = \dfrac{3}{2}$ ,

implying that $p$ can equal either $1$ or $-1$ , respectively. Thus, from the options given, the correct choice is $\boxed{-1}.$

I substituted p for -k, then multiplied numerator and denominator by various $a^k$ until I had positive powers of integers everywhere. And I ended up with

$\frac{27^k + 8^k}{18^k + 12^k}$

It turns out that this only works because 8 × 27 = 12 × 18. The more general formula (found by multiplying num and denom by $(abcd)^k$ ) is

$\frac{a^{-k} + b^{-k}}{c^{-k} + d^{-k}} = \frac{(cd)^k (a^k + b^k)}{(ab)^k (c^k + d^k)}$

Thats so simple.If you take power -1 then all the values reverses and take their L.C.M which will give the valueof 216 same is the L.C.M of upr and lower case which will cutt each other and hence wo got the same value as above.

Here's one way to show that $-1$ is a solution, but one should note that this solution does not show that $p=\pm 1$ only. For the sake of completeness, read Brian's Charlesworth solution.

Suppose we have $A \times B = C \times D$ for positive numbers $A,B,C,D$ , then

$\begin{aligned} \frac {A+B}{C+D} &=& \frac {A+B}{C+D} \times \frac { \frac {1}{AB}}{\frac {1}{CD} } \\ &=& \frac { \frac{A+B}{AB} }{ \frac {C+D}{CD} } \\ &=& \frac { \frac {1}{A} + \frac {1}{B} }{ \frac {1}{C} + \frac {1}{D} } \\ &=& \frac { A^{-1} + B^{-1} }{ C^{-1} + D^{-1} } \\ \end{aligned}$

In this case, $A=8,B=27,C=12,D=18,p=\boxed{-1}$

Let 2^p = x , 3^p = y

Then

(x^3 + y^3)/x y(x + y) = 35/30

(x^2 - x y + y^2)/x y = 7/6

6(x^2) - 13 x y + 6(y^2) = 0

(2 x - 3 y)(3 x - 2 y) = 0

Then

2 x - 3 y = 0 ......................... (1)

Or

3 x - 2 y = 0 ..........................(2)

From Eq (1)

2^(p + 1) = 3^(p + 1)

Then

p + 1 = 0

p = -1

From Eq (2)

2^(p - 1) = 3^(p - 1)

Then

p - 1 = 0

p = 1

A x B =C x D

It is very simple if we take inverse and solve, we get the same as right hand side.

LHS =1/8+1/27////1/12+1/18 =35/216/////30/216 =35/30 =7/6 RHS 35/30 =7/6

its easy guys,mark an crossing product each other and you will see thats thereis -1 thanks