Factorise the following 81x^4-z^4
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Answered by
3
81x^4-z^4
=(9x^2)^2-(z^2)^2
=(9x^2-z^2)(9x^2+z^2)
=(3x)^2-z^2 "
(3x-z)(3x+z)(9x^2+z^2)
=(9x^2)^2-(z^2)^2
=(9x^2-z^2)(9x^2+z^2)
=(3x)^2-z^2 "
(3x-z)(3x+z)(9x^2+z^2)
Answered by
2
81x^4-z^4
=(9^2x^2)^2-(z^2)^2
=(9x)^2-(z^2)^2
={(9x)^2+z^2}{(9x)^2-z^2}
=81x^2+z^2[(9x-z)(9x+z)]
=(9^2x^2)^2-(z^2)^2
=(9x)^2-(z^2)^2
={(9x)^2+z^2}{(9x)^2-z^2}
=81x^2+z^2[(9x-z)(9x+z)]
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