If x=a(b-c),y=b(c-a)and z=c(a-b),then (x/a)^3 +(y/b)^3 +(z/c)^3 =? N.B:^ = cube
D Debasish New member Joined Jan 19, 2012 Messages 13 Jan 23, 2012 #1 If x=a(b-c),y=b(c-a)and z=c(a-b),then (x/a)^3 +(y/b)^3 +(z/c)^3 =? N.B:^ = cube
D Deleted member 4993 Guest Jan 23, 2012 #2 Debasish said: If x=a(b-c),y=b(c-a)and z=c(a-b),then (x/a)^3 +(y/b)^3 +(z/c)^3 =? N.B:^ = cube Click to expand... Do you know how to expand (a-b)3 ?
Debasish said: If x=a(b-c),y=b(c-a)and z=c(a-b),then (x/a)^3 +(y/b)^3 +(z/c)^3 =? N.B:^ = cube Click to expand... Do you know how to expand (a-b)3 ?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jan 23, 2012 #3 Hello, Debasish! If {x = a(b−c)y = b(c−a)z = c(a−b)} find: (xa)3+(yb)3+(zc)3\displaystyle \text{If }\,\begin{Bmatrix}x\:=\:a(b-c) \\ y\:=\:b(c-a) \\ z\:=\:c(a-b)\end{Bmatrix}\;\text{ find: }\:\left(\dfrac{x}{a}\right)^3 +\left(\dfrac{y}{b}\right)^3 +\left(\dfrac{z}{c}\right)^3If ⎩⎪⎨⎪⎧x=a(b−c)y=b(c−a)z=c(a−b)⎭⎪⎬⎪⎫ find: (ax)3+(by)3+(cz)3 Click to expand... We have: .xa = b−c,yb = c−a,zc = a−b\displaystyle \dfrac{x}{a} \:=\:b-c, \quad \dfrac{y}{b} \:=\:c-a, \quad \dfrac{z}{c} \:=\:a-b ax=b−c,by=c−a,cz=a−b Hence: .(xa)3+(yb)3+(zc)3 = (b−c)3+(c−a)3+(a−b)3\displaystyle \left(\dfrac{x}{a}\right)^3 + \left(\dfrac{y}{b}\right)^3 + \left(\dfrac{z}{c}\right)^3 \;=\; (b-c)^3 + (c-a)^3 + (a-b)^3(ax)3+(by)3+(cz)3=(b−c)3+(c−a)3+(a−b)3 . . . . . = b3−3b2c+3bc2−c3+c3−3c2a+3ca2−a3+a3−3a2b+3ab2−b3\displaystyle =\;b^3 - 3b^2c + 3bc^2 - c^3 + c^3 - 3c^2a + 3ca^2 - a^3 + a^3 - 3a^2b + 3ab^2 - b^3=b3−3b2c+3bc2−c3+c3−3c2a+3ca2−a3+a3−3a2b+3ab2−b3 . . . . . = −3b2c+3bc2−3ac2+3a2c−3a2b+3ab2\displaystyle =\;-3b^2c + 3bc^2 - 3ac^2 + 3a^2c - 3a^2b + 3ab^2=−3b2c+3bc2−3ac2+3a2c−3a2b+3ab2 . . . . . = −3[a2b−a2c−ab2+ac2+b2c−bc2]\displaystyle =\;-3\bigg[a^2b - a^2c - ab^2 + ac^2 + b^2c - bc^2\bigg]=−3[a2b−a2c−ab2+ac2+b2c−bc2] . . . . . = −3[a2(b−c)−a(b2−c2)+bc(b−c)]\displaystyle =\;-3\bigg[a^2(b-c) - a(b^2-c^2) + bc(b-c)\bigg]=−3[a2(b−c)−a(b2−c2)+bc(b−c)] . . . . . = −3[a2(b−c)−a(b−c)(b+c)+bc(b−c)]\displaystyle =\;-3\bigg[a^2(b-c) - a(b-c)(b+c) + bc(b-c)\bigg]=−3[a2(b−c)−a(b−c)(b+c)+bc(b−c)] . . . . . = −3(b−c)[a2−a(b+c)+bc]\displaystyle =\;-3(b-c)\bigg[a^2 - a(b+c) + bc\bigg]=−3(b−c)[a2−a(b+c)+bc] . . . . . = −3(b−c)[a2−ab−ac+bc]\displaystyle =\;-3(b-c)\bigg[a^2 - ab - ac + bc\bigg]=−3(b−c)[a2−ab−ac+bc] . . . . . = −3(b−c)[a(a−b)−c(a−b)]\displaystyle =\;-3(b-c)\bigg[a(a-b) - c(a-b)\bigg]=−3(b−c)[a(a−b)−c(a−b)] . . . . . = −3(b−c)(a−b)[a−c]\displaystyle =\;-3(b-c)(a-b)\big[a - c\big]=−3(b−c)(a−b)[a−c] . . . . . = 3(a−b)(b−c)(c−a)\displaystyle =\;\boxed{3(a-b)(b-c)(c-a)}=3(a−b)(b−c)(c−a) . . . . . = 3(zc)(xa)(yb)\displaystyle =\;3\left(\dfrac{z}{c}\right)\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)=3(cz)(ax)(by) . . . . . = 3(xa)(yb)(zc)\displaystyle =\;\boxed{3\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)\left(\dfrac{z}{c}\right)} =3(ax)(by)(cz)
Hello, Debasish! If {x = a(b−c)y = b(c−a)z = c(a−b)} find: (xa)3+(yb)3+(zc)3\displaystyle \text{If }\,\begin{Bmatrix}x\:=\:a(b-c) \\ y\:=\:b(c-a) \\ z\:=\:c(a-b)\end{Bmatrix}\;\text{ find: }\:\left(\dfrac{x}{a}\right)^3 +\left(\dfrac{y}{b}\right)^3 +\left(\dfrac{z}{c}\right)^3If ⎩⎪⎨⎪⎧x=a(b−c)y=b(c−a)z=c(a−b)⎭⎪⎬⎪⎫ find: (ax)3+(by)3+(cz)3 Click to expand... We have: .xa = b−c,yb = c−a,zc = a−b\displaystyle \dfrac{x}{a} \:=\:b-c, \quad \dfrac{y}{b} \:=\:c-a, \quad \dfrac{z}{c} \:=\:a-b ax=b−c,by=c−a,cz=a−b Hence: .(xa)3+(yb)3+(zc)3 = (b−c)3+(c−a)3+(a−b)3\displaystyle \left(\dfrac{x}{a}\right)^3 + \left(\dfrac{y}{b}\right)^3 + \left(\dfrac{z}{c}\right)^3 \;=\; (b-c)^3 + (c-a)^3 + (a-b)^3(ax)3+(by)3+(cz)3=(b−c)3+(c−a)3+(a−b)3 . . . . . = b3−3b2c+3bc2−c3+c3−3c2a+3ca2−a3+a3−3a2b+3ab2−b3\displaystyle =\;b^3 - 3b^2c + 3bc^2 - c^3 + c^3 - 3c^2a + 3ca^2 - a^3 + a^3 - 3a^2b + 3ab^2 - b^3=b3−3b2c+3bc2−c3+c3−3c2a+3ca2−a3+a3−3a2b+3ab2−b3 . . . . . = −3b2c+3bc2−3ac2+3a2c−3a2b+3ab2\displaystyle =\;-3b^2c + 3bc^2 - 3ac^2 + 3a^2c - 3a^2b + 3ab^2=−3b2c+3bc2−3ac2+3a2c−3a2b+3ab2 . . . . . = −3[a2b−a2c−ab2+ac2+b2c−bc2]\displaystyle =\;-3\bigg[a^2b - a^2c - ab^2 + ac^2 + b^2c - bc^2\bigg]=−3[a2b−a2c−ab2+ac2+b2c−bc2] . . . . . = −3[a2(b−c)−a(b2−c2)+bc(b−c)]\displaystyle =\;-3\bigg[a^2(b-c) - a(b^2-c^2) + bc(b-c)\bigg]=−3[a2(b−c)−a(b2−c2)+bc(b−c)] . . . . . = −3[a2(b−c)−a(b−c)(b+c)+bc(b−c)]\displaystyle =\;-3\bigg[a^2(b-c) - a(b-c)(b+c) + bc(b-c)\bigg]=−3[a2(b−c)−a(b−c)(b+c)+bc(b−c)] . . . . . = −3(b−c)[a2−a(b+c)+bc]\displaystyle =\;-3(b-c)\bigg[a^2 - a(b+c) + bc\bigg]=−3(b−c)[a2−a(b+c)+bc] . . . . . = −3(b−c)[a2−ab−ac+bc]\displaystyle =\;-3(b-c)\bigg[a^2 - ab - ac + bc\bigg]=−3(b−c)[a2−ab−ac+bc] . . . . . = −3(b−c)[a(a−b)−c(a−b)]\displaystyle =\;-3(b-c)\bigg[a(a-b) - c(a-b)\bigg]=−3(b−c)[a(a−b)−c(a−b)] . . . . . = −3(b−c)(a−b)[a−c]\displaystyle =\;-3(b-c)(a-b)\big[a - c\big]=−3(b−c)(a−b)[a−c] . . . . . = 3(a−b)(b−c)(c−a)\displaystyle =\;\boxed{3(a-b)(b-c)(c-a)}=3(a−b)(b−c)(c−a) . . . . . = 3(zc)(xa)(yb)\displaystyle =\;3\left(\dfrac{z}{c}\right)\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)=3(cz)(ax)(by) . . . . . = 3(xa)(yb)(zc)\displaystyle =\;\boxed{3\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)\left(\dfrac{z}{c}\right)} =3(ax)(by)(cz)
