Show that (xa-b)a+b (xb-c)b+c (xc-a)c+a =1
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lhs = (x^a-b)^a+b (x^b-c)^b+c (x^c-a)^c+a
= x^(a-b)(a+b) *x^(b-c)(b+c)*x^(c-a)(c+a) here we used (a^m)^n = a^mn
=x^(a²-b²) * x^(b²-c²) *x^(c²-a²) here we used (x+y)(x-y) = x²-y²
=x^(a2-b²+b²-c²+c²-a² )here we used a^m *a^n =a^m+n
=x^0
=1 since x^0=1
=rhs
= x^(a-b)(a+b) *x^(b-c)(b+c)*x^(c-a)(c+a) here we used (a^m)^n = a^mn
=x^(a²-b²) * x^(b²-c²) *x^(c²-a²) here we used (x+y)(x-y) = x²-y²
=x^(a2-b²+b²-c²+c²-a² )here we used a^m *a^n =a^m+n
=x^0
=1 since x^0=1
=rhs
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