JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 17)
The sum of 162th power of the roots of the equation x3 $$-$$ 2x2 + 2x $$-$$ 1 = 0 is ________.
Answer
3
Explanation
x3 $$-$$ 2x2 + 2x $$-$$ 1 = 0
x = 1 satisfying the equation
$$ \therefore $$ x $$-$$ 1 is factor of
x3 $$-$$ 2x2 + 2x $$-$$ 1
= (x $$-$$ 1) (x2 $$-$$ x + 1) = 0
x = 1, $${{1 + i\sqrt 3 } \over 2},{{1 - i\sqrt 3 } \over 2}$$
x = 1, $$-$$ $$\omega$$2, $$-$$$$\omega$$
Sum of 162th power of roots
= (1)162 + ($$-$$$$\omega$$2)162 + ($$-$$$$\omega$$)162
= 1 + ($$\omega$$)324 + ($$\omega$$)162
= 1 + 1 + 1 = 3
x = 1 satisfying the equation
$$ \therefore $$ x $$-$$ 1 is factor of
x3 $$-$$ 2x2 + 2x $$-$$ 1
= (x $$-$$ 1) (x2 $$-$$ x + 1) = 0
x = 1, $${{1 + i\sqrt 3 } \over 2},{{1 - i\sqrt 3 } \over 2}$$
x = 1, $$-$$ $$\omega$$2, $$-$$$$\omega$$
Sum of 162th power of roots
= (1)162 + ($$-$$$$\omega$$2)162 + ($$-$$$$\omega$$)162
= 1 + ($$\omega$$)324 + ($$\omega$$)162
= 1 + 1 + 1 = 3
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