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- Thread starter Debasish
- Start date
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Deleted member 4993
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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
Do you know how to expand (a-b)3 ?
Hello, Debasish!
We have: .\(\displaystyle \dfrac{x}{a} \:=\:b-c, \quad \dfrac{y}{b} \:=\:c-a, \quad \dfrac{z}{c} \:=\:a-b \)
Hence: .\(\displaystyle \left(\dfrac{x}{a}\right)^3 + \left(\dfrac{y}{b}\right)^3 + \left(\dfrac{z}{c}\right)^3 \;=\; (b-c)^3 + (c-a)^3 + (a-b)^3\)
. . . . . \(\displaystyle =\;b^3 - 3b^2c + 3bc^2 - c^3 + c^3 - 3c^2a + 3ca^2 - a^3 + a^3 - 3a^2b + 3ab^2 - b^3\)
. . . . . \(\displaystyle =\;-3b^2c + 3bc^2 - 3ac^2 + 3a^2c - 3a^2b + 3ab^2\)
. . . . . \(\displaystyle =\;-3\bigg[a^2b - a^2c - ab^2 + ac^2 + b^2c - bc^2\bigg]\)
. . . . . \(\displaystyle =\;-3\bigg[a^2(b-c) - a(b^2-c^2) + bc(b-c)\bigg]\)
. . . . . \(\displaystyle =\;-3\bigg[a^2(b-c) - a(b-c)(b+c) + bc(b-c)\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)\bigg[a^2 - a(b+c) + bc\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)\bigg[a^2 - ab - ac + bc\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)\bigg[a(a-b) - c(a-b)\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)(a-b)\big[a - c\big]\)
. . . . . \(\displaystyle =\;\boxed{3(a-b)(b-c)(c-a)}\)
. . . . . \(\displaystyle =\;3\left(\dfrac{z}{c}\right)\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)\)
. . . . . \(\displaystyle =\;\boxed{3\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)\left(\dfrac{z}{c}\right)} \)
\(\displaystyle \text{If }\,\begin{Bmatrix}x\:=\:a(b-c) \\ y\:=\:b(c-a) \\ z\:=\:c(a-b)\end{Bmatrix}\;\text{ find: }\:\left(\dfrac{x}{a}\right)^3 +\left(\dfrac{y}{b}\right)^3 +\left(\dfrac{z}{c}\right)^3\)
We have: .\(\displaystyle \dfrac{x}{a} \:=\:b-c, \quad \dfrac{y}{b} \:=\:c-a, \quad \dfrac{z}{c} \:=\:a-b \)
Hence: .\(\displaystyle \left(\dfrac{x}{a}\right)^3 + \left(\dfrac{y}{b}\right)^3 + \left(\dfrac{z}{c}\right)^3 \;=\; (b-c)^3 + (c-a)^3 + (a-b)^3\)
. . . . . \(\displaystyle =\;b^3 - 3b^2c + 3bc^2 - c^3 + c^3 - 3c^2a + 3ca^2 - a^3 + a^3 - 3a^2b + 3ab^2 - b^3\)
. . . . . \(\displaystyle =\;-3b^2c + 3bc^2 - 3ac^2 + 3a^2c - 3a^2b + 3ab^2\)
. . . . . \(\displaystyle =\;-3\bigg[a^2b - a^2c - ab^2 + ac^2 + b^2c - bc^2\bigg]\)
. . . . . \(\displaystyle =\;-3\bigg[a^2(b-c) - a(b^2-c^2) + bc(b-c)\bigg]\)
. . . . . \(\displaystyle =\;-3\bigg[a^2(b-c) - a(b-c)(b+c) + bc(b-c)\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)\bigg[a^2 - a(b+c) + bc\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)\bigg[a^2 - ab - ac + bc\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)\bigg[a(a-b) - c(a-b)\bigg]\)
. . . . . \(\displaystyle =\;-3(b-c)(a-b)\big[a - c\big]\)
. . . . . \(\displaystyle =\;\boxed{3(a-b)(b-c)(c-a)}\)
. . . . . \(\displaystyle =\;3\left(\dfrac{z}{c}\right)\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)\)
. . . . . \(\displaystyle =\;\boxed{3\left(\dfrac{x}{a}\right)\left(\dfrac{y}{b}\right)\left(\dfrac{z}{c}\right)} \)