JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 21)
If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value of a4 + b4 + c4 is equal to ______________.
Answer
13
Explanation
(a + b + c)2 = 1
$$ \Rightarrow $$ a2 + b2 + c2 + 2(ab + bc + ca) = 1
$$ \Rightarrow $$ a2 + b2 + c2 = – 3 ….(i)
$$ \Rightarrow $$ ab + bc + ca = 2 ….(ii)
Squaring of equation (ii),
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 + 2(ab2c + bc2a + ca2b) = 4
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 + 2abc(a + b + c) = 4
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 + 6 = 4
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 = – 2 ….(iii)
Squaring of equation (i),
$$ \Rightarrow $$ a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 9
$$ \Rightarrow $$ a4 + b4 + c4 – 4 = 9
$$ \Rightarrow $$ a4 + b4 + c4 = 13
$$ \Rightarrow $$ a2 + b2 + c2 + 2(ab + bc + ca) = 1
$$ \Rightarrow $$ a2 + b2 + c2 = – 3 ….(i)
$$ \Rightarrow $$ ab + bc + ca = 2 ….(ii)
Squaring of equation (ii),
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 + 2(ab2c + bc2a + ca2b) = 4
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 + 2abc(a + b + c) = 4
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 + 6 = 4
$$ \Rightarrow $$ a2b2 + b2c2 + c2a2 = – 2 ….(iii)
Squaring of equation (i),
$$ \Rightarrow $$ a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 9
$$ \Rightarrow $$ a4 + b4 + c4 – 4 = 9
$$ \Rightarrow $$ a4 + b4 + c4 = 13
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