Question

if(a+1 divided by a)^2=3 and a not equal to 0;show that a^3+1 divided by a^3 using expansions

Answer 1

Answer:

0 & ±6√3

Step-by-step explanation:

$\bullet\ (a+\frac{1}{a})^2=3 \\ \\ \bullet\ a+\frac{1}{a}=\pm \sqrt{3}$

⇒ Taking a + 1/a = √3,

$\bullet\ (a+\frac{1}{a})^3=(\sqrt{3})^3 \\ \\ \bullet\ a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=3\sqrt{3} \\ \\ \bullet\ a^3+\frac{1}{a^3}\pm3\sqrt{3}=3\sqrt{3} \\ \\ \bullet\ a^3+\frac{1}{a^3}=3\sqrt{3}\mp3\sqrt{3} \\ \\ \bullet\ a^3+\frac{1}{a^3}=\bold{0}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{OR}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^3+\frac{1}{a^3}=\bold{6\sqrt{3}}$

⇒ Taking a + 1/a = -√3,

$\bullet\ (a+\frac{1}{a})^3=(-\sqrt{3})^3 \\ \\ \bullet\ a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=-3\sqrt{3} \\ \\ \bullet\ a^3+\frac{1}{a^3}\pm3\sqrt{3}=-3\sqrt{3} \\ \\ \bullet\ a^3+\frac{1}{a^3}=-3\sqrt{3}\mp3\sqrt{3} \\ \\ \bullet\ a^3+\frac{1}{a^3}=\bold{-6\sqrt{3}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{OR}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^3+\frac{1}{a^3}=\bold{0}$

⇒ Thus the answer is 0 and ±6√3.