A+b+c=6., a3+b3+c3-3abc=126
ab+bc+ca=?
Answers
Application of Algebraic identity
Answer: when a + b + c = 6 and a³+b³+c³-3abc = 126 , value of ab + bc + ca = 5
Explanation:
Given that a + b + c = 6 and a³+b³+c³-3abc=126
need to determine ab + bc + ca .
lets first use algebraic identity of a³+b³+c³-3abc which is as follows
a³+b³+c³-3abc = (a + b + c ) ( a² + b² + c² - ab - bc - ca )
on substituting a³+b³+c³-3abc = 126 and (a + b + c ) = 6 in above identity , we have
126 = (6) ( a² + b² + c² - ab - bc - ca)
=> 126/6 = a² + b² + c² - ab - bc - ca
=> 21 = a² + b² + c² - ab - bc - ca
=> ab + bc + ca = a² + b² + c² - 21 ----------------eq(1)
Now we do not have value of a² + b² + c² but we have a+b+c=6 .
so we will be using following algebraic identity to get a² + b² + c² that is
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
On substituting a + b + c = 6 in above identity we get
(6)² = a² + b² + c² + 2ab + 2bc + 2ca
=> 36 -2ab - 2bc - 2ca = a² + b² + c²
=> a² + b² + c² = 36 -2ab - 2bc - 2ca
On substituting a² + b² + c² = 36 -2ab - 2bc - 2ca in eq(1) we get
ab + bc + ca = 36 -2ab - 2bc - 2ca - 21
=> ab + 2ab + bc + 2bc + ca + 2ca = 36 - 21
=> 3ab + 3bc + 3ca = 15
=> 3 ( ab + bc + ca ) = 15
=> ab + bc + ca = 15/3 = 5
=> ab + bc + ca = 5
Hence when a + b + c = 6 and a³+b³+c³-3abc = 126 , value of ab + bc + ca = 5.
#answerwithquality
#BAL
Step-by-step explanation:
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