Question 1031967
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{{{(2x+5)^4+(2x+1)^4=82}}} [Hint: Put (2x+3)=y]

Using the hint 2x+3=y, then 2x=y-3.  
Substitute y-3 for 2x:

{{{(y-3+5)^4+(y-3+1)^4=82}}}

{{{(y+2)^4+(y-2)^4=82}}}

Multiply that out and collect terms:

{{{2y^4+48y^2+32=82}}}

{{{2y^4+48y^2-50=0}}}

Divide through by 2

{{{y^4+24y^2-25=0}}}

{{{(y^2-1)(y^2+25)=0}}}

{{{(y-1)(y+1)(y^2-25i^2)=0}}}

{{{(y-1)(y+1)(y-5i)(y+5i)=0}}}

y=1, y=-1, y=5i, y=-5i

Since 2x+3=y

2x+3=1, 2x+3=-1, 2x+3=5i, 2x+3=-5i

2x=-2, 2x=-4, 2x=-3+5i, 2x=-3-5i

x=-1, x=-2, x=-3/2+5/2i, x=-3/2-5/2i

Edwin</pre>