PROVE THAT....circle marking....
plz i need determinant
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3
Answer:
Step-by-step explanation:
It will be like
0[(x-a)(x-c)-0(x-b)] - (x+a)[(x-a)0-(x+c)(x-b)] + (x+b)[(x-a)(x-c)-(0)(x-b)]
= -(x+a)[-(x+c)(x-b)] - (x+b)[(x-a)(x-c)]
= (x+a)*[(x+c)(x-b)] - (x+b)*[(x-a)(x-c)]
= 0
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16
see solution in image ............
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