If....

#a/b + b/a = 10#

And... #a^2/b^2 + b^2/a^2 = 98#

Find... #a^3/b^3 + b^3/a^3#

Also can anyone find a formula to generate higher powers??

Question

If....

#a/b + b/a = 10#

And... #a^2/b^2 + b^2/a^2 = 98#

Find... #a^3/b^3 + b^3/a^3#

Also can anyone find a formula to generate higher powers??

Abigail Allen · Accepted Answer

#970#

.

#a/b+b/a=10#

#(a^2+b^2)/(ab)=10#

#a^2+b^2=10ab#

#a^2/b^2+b^2/a^2=98#

#(a^4+b^4)/(a^2b^2)=98#

#a^4+b^4=98a^2b^2#

#a^3/b^3+b^3/a^3=(a^6+b^6)/(a^3b^3)=((a^2+b^2)(a^4-a^2b^2+b^4))/(a^3b^3)=#

#(10ab(98a^2b^2-a^2b^2))/(a^3b^3)=(10ab(97a^2b^2))/(a^3b^3)=(970a^3b^3)/(a^3b^3)=970#

Genesis Allen · Answer

undefined To find $ \frac{a^3}{b^3} + \frac{b^3}{a^3} $, we can use the identity $ \left(\frac{a}{b} + \frac{b}{a}\right)^3 = \frac{a^3}{b^3} + 3\left(\frac{a}{b} + \frac{b}{a}\right) $. Since we know $ \frac{a}{b} + \frac{b}{a} = 10 $, we can substitute it into the identity:

$$ \left(\frac{a}{b} + \frac{b}{a}\right)^3 = 10^3 = 1000 $$

$$ \frac{a^3}{b^3} + 3 \times 10 = 1000 $$

$$ \frac{a^3}{b^3} = 1000 - 30 $$

$$ \frac{a^3}{b^3} = 970 $$

So, $ \frac{a^3}{b^3} + \frac{b^3}{a^3} = 970 + \frac{b^3}{a^3} $.

We also know $ \frac{a^2}{b^2} + \frac{b^2}{a^2} = 98 $. We can use this to find $ \frac{b^3}{a^3} $:

$$ \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right)^2 = 98^2 = 9604 $$

$$ \frac{a^4}{b^4} + 2 = 9604 $$

$$ \frac{a^4}{b^4} = 9604 - 2 $$

$$ \frac{a^4}{b^4} = 9602 $$

$$ \left(\frac{a^4}{b^4}\right) \times \left(\frac{b^3}{a^3}\right) = \frac{a^3}{b^3} $$

$$ 9602 \times \left(\frac{b^3}{a^3}\right) = 970 $$

$$ \frac{b^3}{a^3} = \frac{970}{9602} $$

$$ \frac{b^3}{a^3} = \frac{485}{4801} $$

So, $ \frac{a^3}{b^3} + \frac{b^3}{a^3} = 970 + \frac{485}{4801} $.

As for finding a formula to generate higher powers, it's possible to generalize the pattern observed in the equations provided, but a specific formula would depend on the nature of the pattern observed and may require further analysis.