Question

(16) When 18 an equation called an Identity. Poore that
the trigonometrie Identity 1+tan²A = Sec²A​

Answer 1

Answer:

$Identity:x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x+y+z=1,xy+yz+zx=−1,xyz=−1puttingvalueswegetx3+y3+z3−3(−1)=(1)(x2+y2+z2+1)x3+y3+z3+3=x2+y2+z2+1x3+y3+z3=x2+y2+z2+1−3x3+y3+z3=x2+y2+z2−2....(1)now(x+y+z)2=x2+y2+z2+2(xy+yz+xz)(1)2=x2+y2+z2+2(−1)1=x2+y2+z2−2x2+y2+z2=2+1x2+y2+z2=3putinequation(1)x3+y3+z3=3−2x3+y3+z3=1$