Question

Show that ab2-b2c+bc2-ac2+a2c-a2b=(a-b)(b-c)(c-a)

Answer 1

SOLUTION

TO PROVE

$\sf{a {b}^{2} - {b}^{2} c + b {c}^{2} - a {c}^{2} + {a}^{2}c - {a}^{2}b = (a - b)(b - c)(c - a) }$

PROOF

RHS

$\sf{(a - b)(b - c)(c - a)}$

$= \sf{(ab - ac - {b}^{2} + bc)(c - a)}$

$= \sf{abc - {a}^{2}b - a {c}^{2} + {a}^{2}c - {b}^{2} c+a {b}^{2} + b {c}^{2} - abc }$

$= \sf{ - {a}^{2}b - a {c}^{2} + {a}^{2}c - {b}^{2} c+a {b}^{2} + b {c}^{2} }$

$= \sf{a {b}^{2} - {b}^{2} c + b {c}^{2} - a {c}^{2} + {a}^{2}c - {a}^{2}b}$

= LHS

Hence proved

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Answer 2

We have

• $\sf{ab^2-a^2b-ac^2+a^2c-b^2c+bc^2}$ in L.HS

• $\sf{\left(a-b\right)\left(b-c\right)\left(c-a\right)}$ in R.H.S.

Change the positions i.e L.H.S To R.H.S and R.H.S To L.H.S

• Note : Changing positions doesn't affect answer

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$\sf{\left(a-b\right)\left(b-c\right)\left(c-a\right)\:\:\:\:\:\:\:\bf{(R.H.S)}}$

$\sf{=(ab+a\left(-c\right)+\left(-b\right)b+\left(-b\right)\left(-c\right))\left(c-a\right)}$

$\sf{=(ab-ac-bb+bc)\left(c-a\right)}$

$\sf{=(ab-ac-b^2+bc)\left(c-a\right)}$

$\sf{=abc+ab\left(-a\right)+\left(-ac\right)c+\left(-ac\right)\left(-a\right)+\left(-b^2\right)c+\left(-b^2\right)\left(-a\right)+bcc+bc\left(-a\right)}$

$\sf{=abc-aab-acc+aac-b^2c+ab^2+bcc-abc}$

$\sf{=ab^2-aab+abc-acc+aac-abc-b^2c+bcc}$

$\sf{=ab^2-aab-acc+aac-b^2c+bcc}$

$\bf{=ab^2-a^2b-ac^2+a^2c-b^2c+bc^2\:\:\:\:\:\:\:\:\:(L.H.S)}$

Hence Proved !!