Advertisements
Advertisements
Question
Prove that : x2+ y2 + z2 - xy - yz - zx is always positive.
Solution
x2 + y2 + z2 - xy - yz - zx
= 2(x2 + y2 + z2 - xy - yz - zx)
= 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx
= x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx
= (x2 + y2 - 2xy) + (z2 + x2 - 2zx) + (y2 + z2 - 2yz)
= (x - y)2 + (z - x)2 + (y - z)2
Since square of any number is positive, the given equation is always positive.
APPEARS IN
RELATED QUESTIONS
Simplify : ( x + 6 )( x + 4 )( x - 2 )
Simplify : ( x + 6 )( x - 4 )( x - 2 )
Simplify using following identity : `( a +- b )(a^2 +- ab + b^2) = a^3 +- b^3`
( 2x + 3y )( 4x2 + 6xy + 9y2 )
Simplify using following identity : `( a +- b )(a^2 +- ab + b^2) = a^3 +- b^3`
`( 3x - 5/x )( 9x^2 + 15 + 25/x^2)`
Find : (a + b)(a + b)
Find : (a + b)(a + b)(a + b)
If x = 3 + 2√2, find :
(i) `1/x`
(ii) `x - 1/x`
(iii) `( x - 1/x )^3`
(iv) `x^3 - 1/x^3`
Using suitable identity, evaluate (97)3
Simplify :
`[(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3]/[(x - y)^3 + (y - z)^3 + (z - x)^3]`
Evaluate :
`[0.8 xx 0.8 xx 0.8 + 0.5 xx 0.5 xx 0.5]/[0.8 xx 0.8 - 0.8 xx 0.5 + 0.5 xx .5]`
