Solve (a^(2)(a 2)-2a(a 2) (a 2))/((a-1)^(2)(a 2))

Question

Simplify the expression

- \frac{3 a ^{2}}{a ^{2} - 2 a + 1}

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a = 0, a = 1

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a \in \emptyset

Evaluate

\frac{a ^{2} ( a \times 2 ) - 2 a ( a \times 2 ) ( a \times 2 )}{( a - 1 ) ^{2} ( a \times 2 )}

To find the roots of the expression,set the expression equal to 0

\frac{a ^{2} ( a \times 2 ) - 2 a ( a \times 2 ) ( a \times 2 )}{( a - 1 ) ^{2} ( a \times 2 )} = 0

Find the domain

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\frac{a ^{2} ( a \times 2 ) - 2 a ( a \times 2 ) ( a \times 2 )}{( a - 1 ) ^{2} ( a \times 2 )} = 0, a \in (- \infty, 0) \cup (0, 1) \cup (1, + \infty)

Calculate

\frac{a ^{2} ( a \times 2 ) - 2 a ( a \times 2 ) ( a \times 2 )}{( a - 1 ) ^{2} ( a \times 2 )} = 0

Use the commutative property to reorder the terms

\frac{a ^{2} \times 2 a - 2 a ( a \times 2 ) ( a \times 2 )}{( a - 1 ) ^{2} ( a \times 2 )} = 0

Use the commutative property to reorder the terms

\frac{a ^{2} \times 2 a - 2 a \times 2 a ( a \times 2 )}{( a - 1 ) ^{2} ( a \times 2 )} = 0

Use the commutative property to reorder the terms

\frac{a ^{2} \times 2 a - 2 a \times 2 a \times 2 a}{( a - 1 ) ^{2} ( a \times 2 )} = 0

Use the commutative property to reorder the terms

\frac{a ^{2} \times 2 a - 2 a \times 2 a \times 2 a}{( a - 1 ) ^{2} \times 2 a} = 0

Multiply the terms

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\frac{2 a ^{3} - 2 a \times 2 a \times 2 a}{( a - 1 ) ^{2} \times 2 a} = 0

Multiply

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\frac{2 a ^{3} - 2 ^{3} a ^{3}}{( a - 1 ) ^{2} \times 2 a} = 0

Use the commutative property to reorder the terms

\frac{2 a ^{3} - 2 ^{3} a ^{3}}{2 a ( a - 1 ) ^{2}} = 0

Subtract the terms

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\frac{- 6 a ^{3}}{2 a ( a - 1 ) ^{2}} = 0

Divide the terms

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- \frac{3 a ^{2}}{( a - 1 ) ^{2}} = 0

Rewrite the expression

\frac{- 3 a ^{2}}{( a - 1 ) ^{2}} = 0

Cross multiply

- 3 a^{2} = (a - 1)^{2} \times 0

Simplify the equation

- 3 a^{2} = 0

Change the signs on both sides of the equation

3 a^{2} = 0

Rewrite the expression

a^{2} = 0

The only way a power can be 0 is when the base equals 0

a = 0

Check if the solution is in the defined range

a = 0, a \in (- \infty, 0) \cup (0, 1) \cup (1, + \infty)

Solution

a \in \emptyset

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