(16) When 18 an equation called an Identity. Poore thatthe trigonometrie Identity 1+tan²A = Sec²A About the author Liliana
Answer: [tex]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+1×3+y3+z3=x2+y2+z2+1−3×3+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+z2−2×2+y2+z2=2+1×2+y2+z2=3putinequation(1)x3+y3+z3=3−2×3+y3+z3=1 [/tex] Reply
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Answer:
[tex]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+1×3+y3+z3=x2+y2+z2+1−3×3+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+z2−2×2+y2+z2=2+1×2+y2+z2=3putinequation(1)x3+y3+z3=3−2×3+y3+z3=1 [/tex]