Find all solutions to
3√x+57−3√x−57=3√6.
3√x+57−3√x−57=3√6(3√x+57−3√x−57)3=6(x+57)−33√(x+57)2(x−57)+33√(x+57)(x−57)2−(x−57)=6114−33√(x+57)2(x−57)+33√(x+57)(x−57)2=6108−33√(x+57)2(x−57)+33√(x+57)(x−57)2=0|:336−3√(x+57)2(x−57)+3√(x+57)(x−57)2=036−3√(x2−572)(x+57)+3√(x2−572)(x−57)=036−3√x2−572(3√x+57−3√x−57)⏟=3√6=03√x2−5723√6=36(x2−572)⋅6=363x2−572=7776x2=7776+3249x2=11025x=±105
check x=105:
3√105+57−3√105−57?=3√63√162−3√48?=1.817120592835.45136177850−3.63424118566?=1.817120592831.81712059283=1.81712059283✓
check x=−105:
3√−105+57−3√−105−57?=3√63√−48−3√−162?=1.81712059283−3.63424118566−(−5.45136177850)?=1.81712059283−3.63424118566+5.45136177850?=1.817120592831.81712059283=1.81712059283✓
WolframAlpha shows a solution of x = 105
Maybe a substitution could solve this??? [ But.....I'm not sure ]
Anyone know an algebraic solution ????
Find all solutions to
3√x+57−3√x−57=3√6.
3√x+57−3√x−57=3√6(3√x+57−3√x−57)3=6(x+57)−33√(x+57)2(x−57)+33√(x+57)(x−57)2−(x−57)=6114−33√(x+57)2(x−57)+33√(x+57)(x−57)2=6108−33√(x+57)2(x−57)+33√(x+57)(x−57)2=0|:336−3√(x+57)2(x−57)+3√(x+57)(x−57)2=036−3√(x2−572)(x+57)+3√(x2−572)(x−57)=036−3√x2−572(3√x+57−3√x−57)⏟=3√6=03√x2−5723√6=36(x2−572)⋅6=363x2−572=7776x2=7776+3249x2=11025x=±105
check x=105:
3√105+57−3√105−57?=3√63√162−3√48?=1.817120592835.45136177850−3.63424118566?=1.817120592831.81712059283=1.81712059283✓
check x=−105:
3√−105+57−3√−105−57?=3√63√−48−3√−162?=1.81712059283−3.63424118566−(−5.45136177850)?=1.81712059283−3.63424118566+5.45136177850?=1.817120592831.81712059283=1.81712059283✓