Question

evaluate (1^0+2^0+3^0)(x^+y^0)​

Answer 1

⇒(1⁰+2⁰+3⁰)(x⁰+y⁰)

⇒(1+1+1)(1+1)                    because [n⁰=1] where n is any number

⇒3+2

⇒5

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