Question 986866
Let {{{z = (x+3)^(1/2)^""}}}


If we cube both sides of the equation above, we get {{{z^3 = ((x+3)^(1/2)^"")^3 = (x+3)^(3/2)}}}


This means the original expression {{{(x+3)^(1/2)^""-(x+3)^(3/2)}}} is equivalent to {{{z-z^3}}}


we can factor out a 'z' to get {{{z-z^3=z(1-z^2)}}}


Then use the difference of squares law {{{z(1-z^2)=z(1-z)(1+z)}}}


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So, {{{z-z^3=z(1-z)(1+z)}}}


Now replace each 'z' with {{{(x+3)^(1/2)^""}}}



{{{z-z^3=z(1-z)(1+z)}}}


{{{(x+3)^(1/2)^""-((x+3)^(1/2)^"")^3=(x+3)^(1/2)^""(1-(x+3)^(1/2)^"")(1+(x+3)^(1/2)^"")}}}


{{{(x+3)^(1/2)^""-(x+3)^(3/2)^""=(x+3)^(1/2)^""(1-(x+3)^(1/2)^"")(1+(x+3)^(1/2)^"")}}}


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Yes it's very ugly, but it's a factored form of the original expression


The fully factored form is {{{(x+3)^(1/2)^""(1-(x+3)^(1/2)^"")(1+(x+3)^(1/2)^"")}}}