Question

Factorise : 9x2 + y2 + z2 – 6xy + 2yz - 6zx and hence find its value when x = 1, y = -2 and z = 1​

Answer 1

Answer:

${9x}^{2} + {y}^{2} + {z}^{2} - 6xy + 2yz + - 6zx \\ {(9 \times (1}^{2}) + ( { - 2}^{2} ) + {(1}^{2} ) - (6 \times 1 \times - 2) + (2 \times - 2 \times 1) - (6 \times 1 \times 1) \\ 9 + 4 + 1 - ( - 12) + ( - 4) - 6 \\ 9 + 4 + 1 + 12 + - 4 - 6 \\ 26 - 10 = 16$

Answer 2

Answer:

9x2 + y2 + z2 - 6xy + 2yz - 6zx

= (3x)2 + y2 + z2 - 2 3x y + 2yz - 2 z 3x

= (3x - y - z)2{Using (a-b-c)2

=a2+b2+c2-2ab+2bc-2ca}When x=1,y=2 and z

=-1,9x2 + y2 + z2 - 6xy + 2yz - 6zx

= (3x - y - z)2= [3(1) - 2 - (-1)]2

= (3 - 2 + 1)2 = 22 = 4

By XxitzSehzadixX06