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Posted by Murlee Manohar 7 years, 1 month ago
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Posted by Tan Jaat 7 years, 1 month ago
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Naveen Sharma 7 years, 1 month ago
{tex}3x^2 + 5x - 8 = 0 \\ 3x^2+8x-3x-8 = 0 \\ x(3x+8)-1(3x+8) = 0 \\ (x-1)(3x+8) =0\\ x = 1 , {-8\over 3} {/tex}
Posted by Sabeel Siddiqui 7 years, 1 month ago
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Posted by Sahil Bhandari 7 years, 2 months ago
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Posted by Abhisek Jaiswal 7 years, 2 months ago
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Posted by Sanjay Singh 7 years, 2 months ago
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Posted by Smaty Nickke Singh 7 years, 2 months ago
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Payal Singh 7 years, 2 months ago
Given : {tex}x^2+{1\over x^2}={82\over 9}{/tex}
To find : {tex}x^3-{1\over x^3}{/tex}
Solution:
We know,
{tex}(x-{1\over x})^2= x^2+{1\over x^2}-2{/tex}
{tex}=> (x-{1\over x})^2={82\over 9}-2{/tex}
{tex}=> (x-{1\over x})^2={64\over 9}{/tex}
{tex}=> (x-{1\over x})={8\over 3}{/tex}
Also,
{tex}=> (x^3-{1\over x^3})= (x-{1\over x})(x^2+{1\over x^2}+1){/tex}
{tex}=> (x^3-{1\over x^3})= ({8\over 3})({82\over 9}+1){/tex}
{tex}=> (x^3-{1\over x^3})= ({8\over 3})({91\over 9}) {/tex}
{tex}=> (x^3-{1\over x^3})= {728\over 27}{/tex}
Posted by Dheeraj Ojha 7 years, 2 months ago
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Posted by Abhishek Vishwakarma 7 years, 3 months ago
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Posted by Ricky Ponting 7 years, 3 months ago
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Kritika Trehan 7 years ago
sin(A+B)=sin A cos B + cos A sin B
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