Question

$$\frac { \frac { x ^ { 2 } ( b - c ) } { 2 ^ { 6 } ( a - c ) } \div ( \frac { x ^ { 2 } } { x ^ { 4 } } ) ^ { 2 } = 1 } { }$$

Answer

$$h=1/(99*t^2*a*x^(a+b-(b+a)*c)*(b-c)*(a-c))$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[99that{x}^{a}(b-c){x}^{b}\times \frac{a-c}{{({x}^{b+a})}^{c}}=1\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[99that{x}^{a}(b-c){x}^{b}\times \frac{a-c}{{x}^{(b+a)c}}=1\]
Regroup terms.
\[99ttha{x}^{a}\times \frac{{x}^{b}}{{x}^{(b+a)c}}(b-c)(a-c)=1\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[99{t}^{1+1}ha{x}^{a+b-(b+a)c}(b-c)(a-c)=1\]
Simplify  \(1+1\)  to  \(2\).
\[99{t}^{2}ha{x}^{a+b-(b+a)c}(b-c)(a-c)=1\]
Divide both sides by \(99\).
\[{t}^{2}ha{x}^{a+b-(b+a)c}(b-c)(a-c)=\frac{1}{99}\]
Divide both sides by \({t}^{2}\).
\[ha{x}^{a+b-(b+a)c}(b-c)(a-c)=\frac{\frac{1}{99}}{{t}^{2}}\]
Simplify  \(\frac{\frac{1}{99}}{{t}^{2}}\)  to  \(\frac{1}{99{t}^{2}}\).
\[ha{x}^{a+b-(b+a)c}(b-c)(a-c)=\frac{1}{99{t}^{2}}\]
Divide both sides by \(a\).
\[h{x}^{a+b-(b+a)c}(b-c)(a-c)=\frac{\frac{1}{99{t}^{2}}}{a}\]
Simplify  \(\frac{\frac{1}{99{t}^{2}}}{a}\)  to  \(\frac{1}{99{t}^{2}a}\).
\[h{x}^{a+b-(b+a)c}(b-c)(a-c)=\frac{1}{99{t}^{2}a}\]
Divide both sides by \({x}^{a+b-(b+a)c}\).
\[h(b-c)(a-c)=\frac{\frac{1}{99{t}^{2}a}}{{x}^{a+b-(b+a)c}}\]
Simplify  \(\frac{\frac{1}{99{t}^{2}a}}{{x}^{a+b-(b+a)c}}\)  to  \(\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}}\).
\[h(b-c)(a-c)=\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}}\]
Divide both sides by \(b-c\).
\[h(a-c)=\frac{\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}}}{b-c}\]
Simplify  \(\frac{\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}}}{b-c}\)  to  \(\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}(b-c)}\).
\[h(a-c)=\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}(b-c)}\]
Divide both sides by \(a-c\).
\[h=\frac{\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}(b-c)}}{a-c}\]
Simplify  \(\frac{\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}(b-c)}}{a-c}\)  to  \(\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}(b-c)(a-c)}\).
\[h=\frac{1}{99{t}^{2}a{x}^{a+b-(b+a)c}(b-c)(a-c)}\]