Question 1208462
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Given 144/(2R) = (1/a^2) + (1/b^2) + (1/c^2), solve for R.
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This problem assumes that you have 1 millimeter standard skills of simplifying fractions.



<pre>
Given {{{144/(2R)}}} = {{{1/a^2}}} + {{{1/b^2}}} + {{{1/c^2}}}  is the same as

        {{{72/R}}}   =  {{{1/a^2}}} + {{{1/b^2}}} + {{{1/c^2}}}

        {{{72/R}}} = {{{(b^2*c^2)/(a^2*b^2*c^2)}}} + {{{(a^2*c^2)/(a^2*b^2*c^2)}}} + {{{(a^2*b^2)/(a^2*b^2*c^2)}}} = 

                   = {{{(a^2*b^2 + a^2*c^2 + b^2*c^2)/(a^2*b^2*c^2)}}}.


From here, we get

    R = {{{(72*a^2*b^2*c^2)/(a^2*b^2 + a^2*c^2 + b^2*c^2)}}}.    <U>ANSWER</U>
</pre>

Solved.