If a^{4}+a^{2} b^{2}+b^{4}=8 and a^{2}+a b+b^ | vedclass

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

Question

(D) We know the algebraic identity: $a^{4}+a^{2} b^{2}+b^{4} = (a^{2}+a b+b^{2})(a^{2}-a b+b^{2})$.Given $a^{4}+a^{2} b^{2}+b^{4} = 8$ and $a^{2}+a b+

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We know the algebraic identity: $a^{4}+a^{2} b^{2}+b^{4} = (a^{2}+a b+b^{2})(a^{2}-a b+b^{2})$.Given $a^{4}+a^{2} b^{2}+b^{4} = 8$ and $a^{2}+a b+b^{2} = 4$.Substituting these values into the identity:$8 = 4(a^{2}-a b+b^{2})$$a^{2}-a b+b^{2} = 2$ ........$(1)$We are also given $a^{2}+a b+b^{2} = 4$ ........$(2)$Subtracting equation $(1)$ from equation $(2)$:$(a^{2}+a b+b^{2}) - (a^{2}-a b+b^{2}) = 4 - 2$$a^{2}+a b+b^{2}-a^{2}+a b-b^{2} = 2$$2ab = 2$$ab = 1$.

If $a^{4}+a^{2} b^{2}+b^{4}=8$ and $a^{2}+a b+b^{2}=4,$ then the value of $a b$ is

Similar Questions

$1 \frac{4}{7} + 1 \frac{3}{5} + 1 \frac{1}{3} = ?$

$3 \frac{1}{4} + 2 \frac{1}{2} - 1 \frac{5}{6} = \frac{(?)^{2}}{10} + 1 \frac{5}{12}$

If $\frac{x}{y} = \frac{4}{5}$,then the value of $\left(\frac{4}{7} + \frac{2y - x}{2y + x}\right)$ is

$\frac{1}{\left(2 \frac{1}{3}\right)}+\frac{1}{\left(1 \frac{3}{4}\right)}$ is equal to

Determine the value of $x$. If $x = \frac{(943 + 864)^{2} - (943 - 864)^{2}}{(1886 \times 1728)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module