Math, asked by avanichaudhary200716, 6 months ago

Please answer part= h to i

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Answered by vipashyana1
0

Answer:

h)4a²b-(a²-b)²

=4a²b-(a^4-2a²b+b²)

=4a²b-a^4+2a²b-b²

=-a^4-b²+4a²b+2a²b

=-a^4-b²+6a²b

=a^4+b²-6a²b

i)(xy+3z)²-(xy-3z)(xy-3z)

=(xy+3z)²-(xy-3z)²

=(x²y²+6xyz+3z²)-(x²y²-6xyz+9z²)

=x²y²+6xyz+3z²-x²y²+6xyz-9z²

=x²y²-x²y²+6xyz+6xyz+3z²-9z²

=12xyz-6z²

=6z(2xy-z)

j)[(a+1)(a-1)](a²+1)

=[(a)²-(1)²](a²+1)

=(a²-1)(a²+1)

=(a²)²-(1)²

=a^4-1

k)[(x²-y²)(x²+y²)](x^4+y^4)

=[(x²)²-(y²)²](x^4+y^4)

=(x^4-y^4)(x^4+y^4)

=(x^4)²+(y^4)²

=x^8+y^8

l)(x-1)(x+1)(x²+1)(x^4+1)(x^8+1)

=[(x)²-(1)²](x²+1)(x^4+1)(x^8+1)

=[(x²-1)(x²+1)](x^4+1)(x^8+1)

=[(x²)²-(1)²](x^4+1)(x^8+1)

=[(x^4-1)(x^4+1)](x^8+1)

=[(x^4)²-(1)²](x^8+1)

=[(x^8-1)(x^8+1)]

=[(x^8)²-(1)²]

=(x^16-1)

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