Question

show that (2r - 3s) (2r + 3s) + (3s - 4q)(3s + 4q) + (4q – 2r)(4q + 2r) = 0.​

Answer 1

$\textbf{SOLUTION}$

As we know,

✔️ (a + b)(a-b) = a² - b²

Using this identity, we get

4r² - 9s² + 9s² - 16q² + 16q² - 4r²

(4r² - 4r²) + (9s² - 9s²) + (16q² - 16q²)

0 + 0 + 0

= 0

= R.H.S

✔️ $\huge{Hence\:Proved}$ ✔️

Answer 2

Step-by-step explanation:

(a+b) (a-b)=a²-b²

plus minus get cancel you now then cancel