Question

a cube plus b cube plus c cube minus 3 is equal to a + b + c 1 bracket mein a square + b square + c square minus A B minus b c minus C A​

Answer 1

hey there

here is your answer

we need to prove that :

a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )

PROOF :

now let us take the R.H.S and prove that it is equal to L.H.S

R.H.S :

( a + b + c ) ( a² + b² + c² - ab - bc - ca )    ( MULTIPLYING THE BRACKETS)

=> a ( a² + b² + c² - ab - bc - ca ) + b ( a² + b² + c² - ab - bc - ca ) + c ( a² + b² + c² - ab - bc - ca )

ON MULTIPLYING WE GET

=> a³ + ab² + ac² - a²b - abc - ca² + a²b + b³ + bc² -ab² + b²c - abc + ca² + cb² + c³ - abc - bc² - c²a

( COMBINING ALL LIKE TERMS )

a³ + b³ + c³ + ab² - ab² + ac² - ac² + a²b - a²b + a²c - a²c + bc² - bc² + b²c - b²c - abc - abc -abc

( PLUS AND MINUS OF THE SAME TERMS CANCEL OUT AND WE ARE LEFT WITH )

=> a³ + b³ + c³ - 3abc

L.H.S = R.H.S

HENCE PROVED

hope my answer helps you

please do mark as brainliest

Answer 2

RHS

=> (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)

=> a^3 + ab^2 + ac^2 - a^2 b - abc - a^2 c + a^2 b + b^3 + bc^2 - ab^2 - b^2 c - abc + a^2 c + b^2 c + c^3 - abc - bc^2 - ac^2

=> a^3 + b^3 + c^3 + a^2 b - a^2 b + ab^2 - ab^2 + b^2 c - b^2 c + bc^2 - bc^2 + a^2 c - a^2 c + ac^2 - ac^2 - 3abc

=> a^3 + b^3 + c^3 - 3abc

=> LHS

Hence proved!!!

First I took the RHS, expanded the factors (a + b + c) and (a^2 + b^2 + c^2 - ab - bc - ac), by adding each terms to each other, and then brought the LHS. That's all!

Hope this helps. Plz ask me if you've any doubts.

Thank you. :-))