Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 10005 by konen last updated on 20/Jan/17

(x^(1/(16)) +1)(x^(1/8) +1)(x^(1/4) +1)(x^(1/2) +1)=?

$$\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right)=? \\ $$

Answered by mrW1 last updated on 21/Jan/17

P=(x^(1/(16)) +1)(x^(1/8) +1)(x^(1/4) +1)(x^(1/2) +1) if x=1: P=(1+1)^4 =2^4 =16 if x≠1: (x^(1/(16)) −1)P=(x^(1/(16)) −1)(x^(1/(16)) +1)(x^(1/8) +1)(x^(1/4) +1)(x^(1/2) +1) (x^(1/(16)) −1)P=(x^(1/8) −1)(x^(1/8) +1)(x^(1/4) +1)(x^(1/2) +1) (x^(1/(16)) −1)P=(x^(1/4) −1)(x^(1/4) +1)(x^(1/2) +1) (x^(1/(16)) −1)P=(x^(1/2) −1)(x^(1/2) +1) (x^(1/(16)) −1)P=(x−1) P=((x−1)/(x^(1/(16)) −1))

$${P}=\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right) \\ $$$$ \\ $$$${if}\:{x}=\mathrm{1}: \\ $$$${P}=\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{4}} =\mathrm{2}^{\mathrm{4}} =\mathrm{16} \\ $$$$ \\ $$$${if}\:{x}\neq\mathrm{1}: \\ $$$$\left({x}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right){P}=\left({x}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right) \\ $$$$\left({x}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right){P}=\left({x}^{\frac{\mathrm{1}}{\mathrm{8}}} −\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right) \\ $$$$\left({x}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right){P}=\left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right) \\ $$$$\left({x}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right){P}=\left({x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right) \\ $$$$\left({x}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}\right){P}=\left({x}−\mathrm{1}\right) \\ $$$${P}=\frac{{x}−\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{16}}} −\mathrm{1}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com