[tex]\left(x+\dfrac{1}{x}\right)^3=x^3+3x+\dfrac{3}{x}+\dfrac{1}{x^3}=x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)[/tex]
Zatem
[tex]x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)[/tex]
A więc
[tex]x^3+\dfrac{1}{x^3}=4^3-3\cdot4=64-12=52[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
[tex]\left(x+\dfrac{1}{x}\right)^3=x^3+3x+\dfrac{3}{x}+\dfrac{1}{x^3}=x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)[/tex]
Zatem
[tex]x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)[/tex]
A więc
[tex]x^3+\dfrac{1}{x^3}=4^3-3\cdot4=64-12=52[/tex]