Question 77049
{{{n^3-81n=0}}}
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Notice that both terms on the left side contain the common factor n.  Factor it out to
get:
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{{{n*(n^2-81)=0}}}
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Next notice that the expression within the parentheses is in the form of the difference
of two squares.  This is a common form that can be factored according to the following
rule:
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{{{(a^2 - b^2) = (a+b)*(a-b)}}}
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Applying this rule to {{{(n^2-81)}}} results in the factors {{{(n+9)*(n-9)}}}. Substituting this
into our first factor changes it to:
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{{{n*(n^2-81) = n*(n+9)*(n-9) = 0}}}
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This equation will be true if any of the three factors on the left side equal zero. Therefore,
set each of the factors equal to zero and solve for the corresponding value of n, and the
resulting three values of n will make the equation correct.
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First, let {{{n = 0}}}. No solving is necessary.  This tells you immediately one value of n
that will cause the original equation of {{{n^3-81n=0}}} to become 0 = 0.
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Next, let {{{n+9 =0}}}. Subtract 9 from both sides of the equation, and the resulting
equation is {{{n = -9}}}. This tells you that if you let n = -9 in the original equation,
the original equation of {{{n^3-81n=0}}} will become 0 = 0.
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Finally, let {{{n -9 = 0}}}.  Add 9 to both sides of this equation and the resulting
equation is {{{n = 9}}}.  If you let n = +9 in the original equation, the original equation
of {{{n^3-81n=0}}} will become 0 = 0.
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So the solutions to the original problem are n = 0, n = -9, and n= +9.
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Hope this helps and shows you a way that you can solve for equations that can be factored.