If y=root 3 +root 4.find the value of y cube=1/y cube
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Step-by-step explanation:
Answer:
\frac{1}{x^{2}}+\frac{1}{y^{2}}=194
Step-by-step explanation:
Given\: x=7+4\sqrt{3}\:--(1)
xy = =1\implies y =\frac{1}{x}\:--(2)
y = \frac{1}{7+4\sqrt{3}}
=\frac{7-4\sqrt{3}}{(7+4\sqrt{3})(7-4\sqrt{3})}
=\frac{7-4\sqrt{3}}{7^{2}-\left(4\sqrt{3}\right)^{2}}
=\frac{7-4\sqrt{3}}{49-48}
=7-4\sqrt{3}\: ---(3)
Now,\\\frac{1}{x^{2}}+\frac{1}{y^{2}}
=\frac{1}{\frac{1}{y^{2}}}+x^{2}
=y^{2}+x^{2}\\=(7-4\sqrt{3})^{2}+(7+4\sqrt{3})^{2}\\=2[7^{2}+(4\sqrt{3})^{2}]
/* We know the algebraic identity:
(a-b)²+(a+b)² = 2(a²+b²) */
=2(49+48)\\=2\times 97\\=194
Therefore,
\frac{1}{x^{2}}+\frac{1}{y^{2}}=194
i hope it helps you
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Step-by-step explanation:
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