Question

При 1/a+a=4
Найдите 1/a^3+a^3?

Примет показан и на фото вверху

Answer 1

$\dfrac{1}{a}+a=4 \\\\\Big(\dfrac{1}{a}+a\Big)^{3} =4^{3}\\\\\dfrac{1}{a^{3} } +3\cdot\Big(\dfrac{1}{a}\Big)^{2}\cdot a+3\cdot \dfrac{1}{a}\cdot a^{2}+a^{3}=64\\\\\dfrac{1}{a^{3} } +3\cdot\dfrac{1}{a}+3\cdot a+a^{3}=64\\\\\dfrac{1}{a^{3} } +3\cdot \Big(\dfrac{1}{a}+a\Big)+a^{3}=64\\\\\dfrac{1}{a^{3} } +3\cdot 4+a^{3}=64\\\\\dfrac{1}{a^{3} } +12+a^{3}=64\\\\\boxed{\dfrac{1}{a^{3} }+a^{3}=52}$

Answer 2

$\displaystyle\\\boxed{ x^3+y^3=(x+y)(x^2-xy+y)}}\\\\\left \{ {{(\frac{1}{a} +a)^2=4 ^2 } \atop {\frac{1}{a^3} +a^3}=?} \right.= \left \{ \frac{1}{a^2}+a^2+2\cdot\frac{1}{a}\cdot a =16} \atop {\frac{1}{a^3} +a^3}={{(a+\frac{1}{a})(a^2-a\cdot\frac{1}{a} +\frac{1}{a^2} ) }\right.}= \\\\\\\left \{ {{a^2+\frac{1}{a^2} =14} \atop {\frac{1}{a^3} +a^3}=\underbrace{(a+\frac{1}{a})}_4(\underbrace{a^2+\frac{1}{a^2}}_{14}-1)}} \right. = \frac{1}{a^3} +a^3=4\cdot(14-1)=52$                                        

$Otvet: \boxed{\frac{1}{a^3} +a^3=52}$