find the product (√2a+√3b+√4c)(2a^2+3b^2+4c^2-√6ab-√12bc-√8ca)
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(root2 - root3)(root 2 + root 3)
= (root 2)^2 - (root 3)^2
= 2-3
=-1
integer or rational number
= (root 2)^2 - (root 3)^2
= 2-3
=-1
integer or rational number
Answered by
3
Heya !
Here is yr answer.....
=> (√2a+√3b+√4c) (2a²+3b²+4c²-√6ab-√12bc-√8ca)
It is in the form of ----
[ (a+b+c)(a²+b²+c²-ab-bc-ca) = a³+b³+c³-3xyz]
a = √2a, b = √3b , c = √4c
= (√2a)³ + (√3b)³ + (√4c) ³
Hope it hlpz...
Here is yr answer.....
=> (√2a+√3b+√4c) (2a²+3b²+4c²-√6ab-√12bc-√8ca)
It is in the form of ----
[ (a+b+c)(a²+b²+c²-ab-bc-ca) = a³+b³+c³-3xyz]
a = √2a, b = √3b , c = √4c
= (√2a)³ + (√3b)³ + (√4c) ³
Hope it hlpz...
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