Question 1208462
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{{{144/(2R) = 1/(a^2) + 1/(b^2) + 1/(c^2)}}}


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


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


{{{a^2b^2c^2R(72/R) = a^2b^2c^2R(1/(a^2) + 1/(b^2) + 1/(c^2))}}} Multiply both sides by the LCD a^2*b^2*c^2*R to clear out the fractions.


{{{72a^2b^2c^2 = R*(b^2c^2 + a^2c^2 + a^2b^2)}}} Distribute through a^2*b^2*c^2 on the right hand side. Do <u>not</u> distribute the R since we want to isolate it eventually.


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


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


Verification using WolframAlpha
<a href="https://www.wolframalpha.com/input?i=144%2F%282R%29+%3D+%281%2Fa%5E2%29+%2B+%281%2Fb%5E2%29+%2B+%281%2Fc%5E2%29%2C+solve+for+R">https://www.wolframalpha.com/input?i=144%2F%282R%29+%3D+%281%2Fa%5E2%29+%2B+%281%2Fb%5E2%29+%2B+%281%2Fc%5E2%29%2C+solve+for+R</a>
WolframAlpha decided to factor a^2 from a^2c^2 + a^2b^2.
The order of the terms being multiplied doesn't matter (eg: a^2b^2c^2 is the same as b^2a^2c^2), and neither does the order of the terms being added (eg: b^2c^2 + a^2c^2 is the same as a^2c^2+b^2c^2)


Another verification tool you can use is the CAS mode in GeoGebra.
There are many other tools to pick from.


Edit:
A way to avoid writing so many exponents, we can say
{{{R = (72(abc)^2)/((bc)^2 + (ac)^2 + (ab)^2) }}}
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