Show that :
$(2 r - 3 s) (2 r + 3 s) + (3 s - 4 q) (3 s + 4 q) + (4 q - 2 r) (4 q + 2 r) = 0$

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Solution

$L . H . S = (2 r - 3 s) (2 r + 3 s) + (3 s - 4 q) (3 s + 4 q) + (4 q - 2 r) (4 q + 2 r)$

$= {4 r}^{2} - 9 s^{2} + 9 s^{2} - 16 q^{2} + 16 q^{2} - {4 r}^{2}$ using $(a + b) (a - b) = a^{2} - b^{2}$

$= 0 = R . H . S$

Hence proved.

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Show that :(2r−3s)(2r+3s)+(3s−4q)(3s+4q)+(4q−2r)(4q+2r)=0

L.H.S=(2r−3s)(2r+3s)+(3s−4q)(3s+4q)+(4q−2r)(4q+2r)=4r2−9s2+9s2−16q2+16q2−4r2 using (a+b)(a−b)=a2−b2=0=R.H.SHence proved.

Show that :
$(2 r - 3 s) (2 r + 3 s) + (3 s - 4 q) (3 s + 4 q) + (4 q - 2 r) (4 q + 2 r) = 0$

$L . H . S = (2 r - 3 s) (2 r + 3 s) + (3 s - 4 q) (3 s + 4 q) + (4 q - 2 r) (4 q + 2 r)$

$= {4 r}^{2} - 9 s^{2} + 9 s^{2} - 16 q^{2} + 16 q^{2} - {4 r}^{2}$ using $(a + b) (a - b) = a^{2} - b^{2}$

$= 0 = R . H . S$

Hence proved.