Advertisements
Advertisements
Question
Prove that : x2+ y2 + z2 - xy - yz - zx is always positive.
Advertisements
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 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)`
Simplify using following identity : `( a +- b )(a^2 +- ab + b^2) = a^3 +- b^3`
`(a/3 - 3b)(a^2/9 + ab + 9b^2)`
If a + b = 11 and a2 + b2 = 65; find a3 + b3.
If x + 5y = 10; find the value of x3 + 125y3 + 150xy − 1000.
If a − 2b + 3c = 0; state the value of a3 − 8b3 + 27c3.
Using suitable identity, evaluate (104)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 :
`[1.2 xx 1.2 + 1.2 xx 0.3 + 0.3 xx 0.3 ]/[ 1.2 xx 1.2 xx 1.2 - 0.3 xx 0.3 xx 0.3]`
