Question 980126

if
{{{27x^6+81x^4+81x^2+27=(ax^2+b)^3}}}, then find ab

first factor out {{{27}}}

{{{27(x^6+3x^4+3x^2+1)}}}=..........write {{{3x^4}}} as {{{x^4+2x^4}}} and {{{3x^2}}} as {{{2x^2+x^2}}}

{{{27(x^6+x^4+2x^4+2x^2+x^2+1)}}}=.........group

{{{27((x^6+x^4)+(2x^4+2x^2)+(x^2+1))}}}=factor out 

{{{27(x^4(x^2+1)+2x^2(x^2+1)+(x^2+1))}}}=

{{{27(x^4+2x^2+1)(x^2+1)}}}=

{{{27(x^2+1)^2(x^2+1)}}}=.......write {{{27}}} as {{{3^3}}} 

{{{3^3 (x^2+1)^3}}}=

{{{(3x^2+3)^3}}}

so, {{{a=3}}} and {{{b=3}}}