Question 132660
{{{(1/8)(3x+1)^(3/2) + (1/4)(3x+1)^(1/2) }}} Start with the given expression



{{{(1/8)(3x+1)^(3*(1/2)) + (1/4)(3x+1)^(1/2) }}} Rewrite {{{3/2}}} as {{{3*(1/2)}}}



{{{(1/8)((3x+1)^(1/2))^3 + (1/4)(3x+1)^(1/2) }}} Rewrite the first expression using the identity {{{x^(y*z)=(x^y)^z}}}



{{{(1/8)(sqrt(3x+1))^3 + (1/4)sqrt(3x+1) }}} Now rewrite {{{(3x+1)^(1/2)}}} as {{{sqrt(3x+1)}}}




Let {{{u=sqrt(3x+1)}}}



{{{(1/8)u^3 + (1/4)u }}} Plug in {{{u=(3x+1)^(1/2)}}}



{{{(1/4)u((1/2)u^2 + 1) }}} Factor out the GCF {{{(1/4)u}}}



{{{(1/4)(sqrt(3x+1))((1/2)(sqrt(3x+1))^2 + 1) }}} Now substitute {{{sqrt(3x+1)}}} back in for u



{{{(1/4)(sqrt(3x+1))((1/2)(3x+1) + 1) }}} Square {{{sqrt(3x+1)}}} to get  {{{3x+1}}}



{{{(1/4)(sqrt(3x+1))((1/2)3x+1/2 + 1) }}} Distribute



{{{(1/4)(sqrt(3x+1))((1/2)3x+3/2) }}} Combine like terms



{{{sqrt(3x+1)(1/4)((1/2)3x+3/2) }}} Rearrange the terms



{{{sqrt(3x+1)((1/8)3x+3/8) }}} Distribute



{{{((1/8)3x+3/8)sqrt(3x+1) }}} Rearrange the terms