Question

If x + y + z = 6 and xy + yz + zx = 12, then show that x³ + y³ + z³ = 3xyz​

Answer 1

$\mathcal{\huge{\underline{\underline{\red{SOLUTION}}}}}$

ᏩᏆᏙᎬN :-

If x+y+z=6 and xy+yz+zx=12

ᎢᎾ ᏚᎻᎾᏔ :-

x³ + y³ +z³ =3xyz

ᏢᎡᎾᎾF :-

ᴀᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

➡x³ +y³+z³−3xyz=(x+y+z)(x²+y²+z²−xy−yz−zx)

➡(6){(x+y+z)² −2xy−2yz−2zx−xy−yz−zx}

➡(6){(x+y+z)²−3xy−3yz−3zx}

➡(6){(x+y+z)²−3(xy+yz+zx)}

➡(6)(6² −3×12)

➡(6)(36−36)

➡0

Therefore,

✒ x³ + y³ + z³ = 3xyz

_______________________________

Answer 2

Given in the question..

x + y + z = 6.

xy + yz + zx = 12.

Now, solving step by step..

We know the formula that was:-

=> x³ + y³ + z³ – 3xyz.

=>( x + y + z ) ( x² + y² + z² – xy – yz – zx )..

We know that x + y + z = 6 ( Given )...1⃣

& xy + yz + zx = 12 ( Given )....2⃣

=> ( 6 ) {( x + y + z )²–2xy–2yz–2zx–xy–yz – zx }

=> ( 6 ) {( x + y + z )² – 3xy – 3yz – 3zx }

=> ( 6 ) {( x + y + z )² – 3 ( xy – yz – zx ) }

Now we place the value of 1⃣ & 2⃣ and here. We get.

=> ( 6 ) {( 6² — 3 × 12 }

=> ( 6 ) { 36 — 36 }

=> ( 6 ) 0

=> 0

Therefore, x³ + y³ + z³ = 3xyz..

I hope that this help you..

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