if w is imaginary cube root of unity and ( a+bw+cw^2)^2015 = ( a+ bw^2 + cw) qheee a,b,c are unequal real numbers, then a^2 + b^2 + c^2 -ab - bc -ca is equal to ?

jee advance question ( integer type)

3 Answers

Answer is 1...... See!!.. (a+bw+cw^2)*(a+bw^2+cw) =a^2+b^2+c^2 -ab - bc - ca ?? this is true for all values of a, b, c.. And also we have a relation in them as given in ques.. So we have (a+bw+cw^2)^2016 =a^2+b^2+c^2 -ab - bc - ca Now take modulus on both sides.. As LHS is real its modulus value would be same but also.. Modulus of (a+bw+cw^2) is [([a^2+b^2+c^2 -ab - bc - ca]^1/2)] .. check this by putting the values of w and w^2 as one being (-1+ (3^1/2) i) /2 and other be its conjugate ... After taking modulus we get.. a^2+b^2+c^2 -ab - bc - ca = (a^2+b^2+c^2 -ab - bc - ca) ^ (2016/2)... Now a^2+b^2+c^2 -ab - bc - ca = (a^2+b^2+c^2 -ab - bc - ca) ^1008 .. We get.. two values as 0 and 1 both satisfies this but as a, b, c are not equal we take only one value i. e. 1 ... Hence, answer is 1.. Think about it!!..
- Mukul
- 29 Sep
- 0 Comment
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pls post the answer if you get
- Pranit
- 15 May
- 0 Comment
0
qhee = where *
- Shalinie
- 29 Apr
- 0 Comment
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