document.write( "Question 42061: X^1/m=Y^1/n=Z^1/p, xyz=1 then m+n+p=? \n" ); document.write( "
Algebra.Com's Answer #27193 by Fermat(136)\"\" \"About 
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You have,
\n" ); document.write( "x^(1/m)=y^(1/n)=z^(1/p)
\n" ); document.write( "From,
\n" ); document.write( "x^(1/m)=y^(1/n)
\n" ); document.write( "x = y^(m/n) -------------(1)
\n" ); document.write( "===========
\n" ); document.write( "From,
\n" ); document.write( "y^(1/n)=z^(1/p)
\n" ); document.write( "z = y^(p/n) -------------(2)
\n" ); document.write( "===========
\n" ); document.write( "Using (1) and (2),
\n" ); document.write( "xyz = y^(m/n) * y * y^(p/n) = 1
\n" ); document.write( "which gives,
\n" ); document.write( "y^(m/n + 1 + p/n) = 1
\n" ); document.write( "But if y^k = 1, then k = 0!
\n" ); document.write( "So,
\n" ); document.write( "(m/n + 1 + p/n) = 0
\n" ); document.write( "m/n + n/n + p/n = 0
\n" ); document.write( "(m+n+p)/n = 0
\n" ); document.write( "m+n+p = 0
\n" ); document.write( "========= \n" ); document.write( "
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