How do you simplify #root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3))# ?

2 Answers
Apr 20, 2018

Answer:

#root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) = 6#

Explanation:

Let:

#x = root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3))#

Then:

#x^3 = (135+78sqrt(3))+(135-78sqrt(3))+3(root(3)(135+78sqrt(3))root(3)(135-78sqrt(3)))(root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)))#

#color(white)(x^3) = 270+3(root(3)(135^2-78^2(3)))x#

#color(white)(x^3) = 270+3(root(3)(18225-18252))x#

#color(white)(x^3) = 270+3(root(3)(-27))x#

#color(white)(x^3) = 270-9x#

So:

#x^3+9x-270 = 0#

By the rational roots theorem, any rational roots of this cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-270# and #q# a divisor of the coefficient #1# of the leading term.

So the first few possible rational zeros are:

#+-1, +-2, +-3, +-5, +-6, +-9, +-10,...#

By Descartes' Rule of Signs, we can tell that the cubic has exactly one positive real root and no negative real zeros, so try the possible positive rational zeros.

We find:

#(color(blue)(6))^3+9(color(blue)(6))-270 = 216+54-270 = 0#

So #x=6# is a root and #(x-6)# is a factor:

#x^3+9x-270 = (x-6)(x^2+6x+45)#

The remaining quadratic has no real zeros (as we can see by looking at its discriminant).

The non-real zeros are actually:

#omega root(3)(135+78sqrt(3))+omega^2 root(3)(135-78sqrt(3))#

and:

#omega^2 root(3)(135+78sqrt(3))+omega root(3)(135-78sqrt(3))#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

So:

#root(3)(135+78sqrt(3))+root(3)(135-78sqrt(3)) = 6#

Apr 20, 2018

Answer:

# 6#.

Explanation:

Let us note that,

#(x+sqrt3y)^3=x^3+3sqrt3y^3+3sqrt3xy(x+sqrt3y)#,

#=x^3+3sqrt3y^3+3sqrt3x^2y+9xy^2#.

# rArr (x+sqrt3y)^3=(x^3+9xy^2)+3sqrt3(x^2y+y^3), or, #

#root(3){(x^3+9xy^2)+sqrt3(3x^2y+3y^3)}=x+sqrt3y#.

So, from this we guess :

If #root(3)(135+78sqrt3)=x+sqrt3y#, then,

#(x^3+9xy^2)=135, and, (3x^2y+3y^3)=78, i.e., #

# x^3+9xy^2=x(x^2+9y^2)=135......(1), and, #

# 3y(x^2+y^2)=78, or, y(x^2+y^2)=26......(2)#.

#y(x^2+y^2)=26=1xx26, or, 2xx13#

By Trial and Error Method, # y=1, x=5;" not suitable for "(1)#

# y=2, x=3" satisfy "(1)#.

Therefore, #root(3)(135+78sqrt3)=x+sqrt3y=3+2sqrt3#.

Consequently, #root(3)(135-78sqrt3)=3-2sqrt3#.

Finally, #root(3)(135+78sqrt3)+root(3)(135-78sqrt3)=6#,

as Respected George C. Sir has already derived!