If (a+frac{1}{a})^{2}=3, then a^{3}+frac{1}{a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $(a+\frac{1}{a})^{2}=3$.Taking the square root on both sides,we get $a+\frac{1}{a}=\sqrt{3}$.To find $a^{3}+\frac{1}{a^{3}}$,we use the

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given that $(a+\frac{1}{a})^{2}=3$.Taking the square root on both sides,we get $a+\frac{1}{a}=\sqrt{3}$.To find $a^{3}+\frac{1}{a^{3}}$,we use the algebraic identity $(x+y)^{3} = x^{3} + y^{3} + 3xy(x+y)$.Substituting $x=a$ and $y=\frac{1}{a}$,we get $(a+\frac{1}{a})^{3} = a^{3} + \frac{1}{a^{3}} + 3(a)(\frac{1}{a})(a+\frac{1}{a})$.This simplifies to $(a+\frac{1}{a})^{3} = a^{3} + \frac{1}{a^{3}} + 3(a+\frac{1}{a})$.Substituting the value $a+\frac{1}{a}=\sqrt{3}$ into the equation:$(\sqrt{3})^{3} = a^{3} + \frac{1}{a^{3}} + 3(\sqrt{3})$.$3\sqrt{3} = a^{3} + \frac{1}{a^{3}} + 3\sqrt{3}$.Subtracting $3\sqrt{3}$ from both sides,we get $a^{3} + \frac{1}{a^{3}} = 0$.

If $(a+\frac{1}{a})^{2}=3,$ then $a^{3}+\frac{1}{a^{3}}=?$

Similar Questions

The value of $1001 \div 11 \text{ of } 13$ is

If $x=\sqrt[3]{5}+2,$ then the value of $x^{3}-6 x^{2}+12 x-13$ is

If $x * y = x^{2} + y^{2} - xy$,then the value of $9 * 11$ is

$4 \frac{3}{4} + 2 \frac{1}{8} + 7 \frac{1}{4} + 3 \frac{7}{8} + 11 \frac{12}{13} = ?$

$3 \frac{3}{4} + 4 \frac{2}{5} - 3 \frac{1}{8} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module