Question 87941
<pre>
solve equation, by making appropriate substitution. 

Rule: let U equal the variable part of the second term (without the coefficient)
Then UČ will equal the variable part of the first term (without the coefficient)

a){{{x^(-2)+x^(-1)-56=0}}}

{{{U = x^(-1)}}}, which you see is the variable part of the second term

{{{U^2= (x^(-1))^2 = x^(-2)}}}, which is the variable part of the first term 

{{{x^(-2)+x^(-1)-56=0}}} becomes

{{{U^2+U^2-56=0}}}

{{{(U+8)(U-7)=0}}}
{{{U+8=0}}},  {{{U-7=0}}}
{{{U=-8}}},   {{{U=7}}}

Now substitute back, using {{{U = x^(-1)}}},

{{{x^(-1) = -8}}},  {{{x^(-1)=7}}}

Use the definition of a negative exponent to rewrite those:

{{{1/x = -8}}},  {{{ 1/x = 7}}}

{{{1/x = (-8)/1}}}, {{{1/x = 7/1}}}

Cross multiply:

{{{-8x = 1}}}, {{{7x = 1}}}

{{{x = -1/8}}}, {{{x = 1/7}}}

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b){{{x^(2/3)-x^(1/3)-12=0}}}
{{{U=x^(1/3)}}},  the variable part of the second term
{{{U^2= (x^(1/3))^2 = x^(2/3)}}},  the variable part of the first term 

{{{x^(2/3)-x^(1/3)-12=0}}} becomes

{{{U^2-U-12=0}}}

{{{(U-4)(U+3)=0}}}
{{{U-4=0}}},  {{{U+3=0}}}
{{{U=4}}},   {{{U=-3}}}

Now substitute back, using {{{U = x^(1/3)}}},

{{{x^(1/3) = 4}}},  {{{x^(1/3)=-3}}}

Use the definition of a fractional exponent to rewrite those:

{{{root(3,x) = 4}}},  {{{root(3,x) = -3}}}

Cubing both sides of both equations:

{{{(root(3,x))^3 = 4^3}}}, {{{(root(3,x))^3 = (-3)^3}}}

{{{x = 64}}}, {{{x = -27}}}

Edwin</pre>