Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 106664 by deep last updated on 06/Aug/20

Factorise: x^6 + 64y^6

$$\mathrm{Factorise}:\:\:\:{x}^{\mathrm{6}} \:+\:\mathrm{64}{y}^{\mathrm{6}} \\ $$

Answered by som(math1967) last updated on 06/Aug/20

(x^2 )^3 +(4y^2 )^3 =(x^2 +4y^2 )(x^4 −x^2 .4y^2 +16y^4 ) =(x^2 +4y^2 )(x^4 −4x^2 y^2 +16y^4 )

$$\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} +\left(\mathrm{4y}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4y}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} .\mathrm{4y}^{\mathrm{2}} +\mathrm{16y}^{\mathrm{4}} \right) \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4y}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{4}} −\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{16y}^{\mathrm{4}} \right) \\ $$

Answered by nimnim last updated on 06/Aug/20

(x^2 )^3 +(4y^2 )^3 (x^2 +4y^2 ){(x^2 )^2 −(x^2 )(4y^2 )+(4y^2 )^2 } (x^2 +4y^2 )(x^4 −4x^2 y^2 +16y^4 )

$$\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} +\left(\mathrm{4}{y}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \right)\left\{\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} \right)\left(\mathrm{4}{y}^{\mathrm{2}} \right)+\left(\mathrm{4}{y}^{\mathrm{2}} \right)^{\mathrm{2}} \right\} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{16}{y}^{\mathrm{4}} \right) \\ $$

Answered by abdomathmax last updated on 07/Aug/20

let p(x) =x^6 +64y^6 roots of p(x) p(x)=0 ⇔x^6 =−64y^6 ⇒((x/y))^6 =−2^6 ⇒ ((x/(2y)))^6 =−1 =e^(i(2k+1)π) ⇒(x/(2y)) =e^((i(2k+1)π)/6) ⇒the roots are x_k =2y e^((i(2k+1)π)/6) with k∈[[0,5]] ⇒ p(x) =Π_(k=0) ^5 (x−2ye^((i(2k+1)π)/6) )(factorisation atC(x])

$$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}\right)\:=\mathrm{x}^{\mathrm{6}} \:+\mathrm{64y}^{\mathrm{6}} \:\mathrm{roots}\:\mathrm{of}\:\mathrm{p}\left(\mathrm{x}\right) \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{0}\:\Leftrightarrow\mathrm{x}^{\mathrm{6}} \:=−\mathrm{64y}^{\mathrm{6}} \:\Rightarrow\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{6}} \:=−\mathrm{2}^{\mathrm{6}} \:\Rightarrow \\ $$$$\left(\frac{\mathrm{x}}{\mathrm{2y}}\right)^{\mathrm{6}} \:=−\mathrm{1}\:=\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi} \:\Rightarrow\frac{\mathrm{x}}{\mathrm{2y}}\:=\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{6}}} \:\Rightarrow\mathrm{the}\:\mathrm{roots}\:\mathrm{are} \\ $$$$\mathrm{x}_{\mathrm{k}} =\mathrm{2y}\:\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{6}}} \:\:\mathrm{with}\:\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{5}\right]\right]\:\Rightarrow \\ $$$$\mathrm{p}\left(\mathrm{x}\right)\:=\prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \left(\mathrm{x}−\mathrm{2ye}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{6}}} \right)\left(\mathrm{factorisation}\:\mathrm{atC}\left(\mathrm{x}\right]\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com