If $2 x^2+y^2+8 z^2-2 \sqrt{2} x y+4 \sqrt{2} y z-8 z x=(A x+y+B z)^2$, then the value of $\left(A^2+B^2-A B\right)$ is:
Answer & explanation
Correct answer: option 2
(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
= 2x2 + y2 + 8z2– 2√2xy + 4√2yz – 8zx = (Ax + y + Bz)2
= (√2 x)2 + y2 + (2√2 z)2 – 2 × (√2 x) × y + 2 × y × 2√2z – 2 × 2√2 z × √2 x = (Ax + y + Bz)2
= (√2x –y – 2√2 z)2 = (Ax + y + Bz)2
Comparing on both sides
A = √2
B = – 2√2
Now,
(A2 + B2 – AB)
= [(√2)2 + (–2√2)2 – √2 × (–2√2)]
= [2 + 8 + 4] = 14