Question

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Answer 1

$Q6. (i) (100-1)^{3}\\ = 100^{3} - 1^3 - 3*100*1(100-1)\\= 1000000-1-300*99\\= 1000000-1-29700\\= 970299\\(ii) 102^2\\=(100 + 2) ^2\\=100^2 + 2*100*2 + 2^2\\=10000 + 400+4\\= 10404$$(i) 1^{4} - 3*1^{2} + 2*1 +1 = 1 \\\\(ii) 5* -3^{3} + 5* -3^{2} - 6* -3 + 9 = -63\\\\(iii)3*5^{3} - 16*5{2} - 5*5 + 50 = 0\\\\(iv) -4{^3} + -4^{2} - 10*-4 + 8 = 0\\\\\\Q4.\\(i) x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8zx\\(ii) 9a^{2} + 49b^{2} + c^{2} - 42ab + 14bc - 6ca\\\\\\Q5. (x+y+z)^{2}c = x^{2} +y^{2} +z^{2} + 2xy + 2yz + 2zx\\ = x^{2} +y^{2} +z^{2} + 2(xy + yz + zx)\\ = 40 + 2*12\\ = 40+24\\ = 64$

$(x+y+z)^{2} = 64\\x+y+z = \sqrt{64} \\x+y+z = 8$

$x^{4} +( \frac{1}{x^4}) = 119\\\\(x^2)^2 + (\frac{1}{x^2})^2 + 2 = [(x^2) + (\frac{1}{x^2})]^2\\\\119+2 = [(x^2) + (\frac{1}{x^2})]\\\\121 = [(x^2) + (\frac{1}{x^2})]^2\\\\x^2 + \frac{1}{x^2} = \sqrt{121} \\\\x^2 + \frac{1}{x^2} = 11\\\\x^2 + \frac{1}{x^2} - 2 = [x-\frac{1}{x} ]^2\\\\11 - 2 = [x-\frac{1}{x} ]^2\\\\9 = [x-\frac{1}{x} ]^2\\\\x-\frac{1}{x} = \sqrt{9} \\\\x-\frac{1}{x} = 3 \\\\x-\frac{1}{x} ]^3 = 3^3 = 27\\27 = x^3 - \frac{1}{x^3} - 3x^2* \frac{1}{x} + 3x* \frac{1}{x^2}$

Answer 2

Answer:

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