PROVE

(a^2-b^2)^2 + (2ab)^2 = (a^2+b^2)^2

asked by guest
on Nov 17, 2024 at 4:11 pm



You asked:

Prove or disprove that the equation $\left(a^{2} - b^{2}\right)^{2} + \left(2 a b\right)^{2} = \left(a^{2} + b^{2}\right)^{2}$ is an identity.

MathBot Answer:

The left-hand side of the equation evaluates to:

\[\begin{align*}& \left(a^{2} - b^{2}\right)^{2} + \left(2 a b\right)^{2}\\=& a^{4} + b^{4} - 2 a^{2} b^{2} + \left(2 a b\right)^{2}\\=& a^{4} + b^{4} - 2 a^{2} b^{2} + 4 a^{2} b^{2}\\=& a^{4} + b^{4} + 2 a^{2} b^{2}\\=& a^{4} + 2 a^{2} b^{2} + b^{4}\\\end{align*}\]

The right-hand side of the equation evaluates to:

\[\begin{align*}& \left(a^{2} + b^{2}\right)^{2}\\=& a^{4} + b^{4} + 2 a^{2} b^{2}\\=& a^{4} + 2 a^{2} b^{2} + b^{4}\\\end{align*}\]

Since $a^{4} + 2 a^{2} b^{2} + b^{4} = a^{4} + 2 a^{2} b^{2} + b^{4}$, the equation $\left(a^{2} - b^{2}\right)^{2} + \left(2 a b\right)^{2} = \left(a^{2} + b^{2}\right)^{2}$ is an identity.

+ Submit Question