Question

(12.) Verify that x3 + y2 + 23 - 3xyz
= { (x + y + 2) [(x−y) +(3+2)2 +62–192].​

Answer 1

Correct Question :

Verify that x³ + y³ + z³ - 3xyz = 1/2 (x + y + z) [(x - y)² + (y - z)² + (z - x)²]

Step-by-step explanation :

To verify the above equation, we have to prove that, L.H.S. = R.H.S.

So, let's start it by taking R.H.S.

⇒ 1/2 (x + y + z) [(x - y)² + (y - z)² + (z - x)²]

⇒ 1/2 (x + y + z) (x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2zx)

⇒ 1/2 (x + y + z) (2x² + 2y² + 2z² - 2xy - 2yz - 2zx)

⇒1/2 ×2 (x + y + z) (x² + y² +z² - xy - yz - zx)

⇒ (x + y + z) (x² + y² +z² - xy - yz - zx)

We know that

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

⇒ x³ + y³ +z³ - 3xyz

From above calculations we have R.H.S = x³ + y³ + z³ - 3xyz

If we notice the question then we found that L.H.S. is also x³ + y³ + z³ - 3xyz

x³ + y³ + z³ - 3xyz = x³ + y³ + z³ - 3xyz

Therefore,

L.H.S. = R.H.S.

Hence, proved

Answer 2

Question :--- Verify that x³ + y³ + z³ - 3xyz = (1/2) (x + y + z) [(x−y)² +(y-z)² +(z-x)²].

Solution :---

As we know that ,

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

Lets Assume it as Equation (1) .

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Now, Lets Try to solve (x² + y² + z² - xy - yz - zx) First .

→ (x² + y² + z² - xy - yz - zx)

Multiplying and dividing by 2 [ or we can say that Multiplying by 1 actually as 2/2 = 1 ]

→ 1/2 [ 2* (x² + y² + z² - xy - yz - zx) ]

→ 1/2 [ 2x² + 2y² + 2z² - 2xy -2yz - 2zx ]

→ 1/2 [ x² - 2xy + y² + y² - 2yz + z² + z² - 2zx + x² ]

Now, we know that [ a² -2ab + b² = (a-b)² ]

→ 1/2 [(x - y)² + (y - z)² + (z - x)²] --------- Equation (2) .

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So, Now, putting Value of Equation (2) in Equation (1) , we get,

→ x³ + y³ + z³ - 3xyz = (x + y + z) * { 1/2 [(x - y)² + (y - z)² + (z - x)²] }

→ x³ + y³ + z³ - 3xyz = (1/2) (x + y + z) [(x−y)² +(y-z)² +(z-x)²].

✪✪ Hence Proved ✪✪

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If You want to Prove Equation (1) Formula also, Than Just Multiply the terms inside bracket ..

Let see That one also , For Better Explanation ..

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

Taking RHS side we get,

→ (x + y + z) (x² + y² + z² - xy - yz - zx)

→ [x(x² + y² + z² - xy - yz - zx) + y(x² + y² + z² - xy - yz - zx) + z(x² + y² + z² - xy - yz - zx) ]

→ [ x³ + xy² + xz² -x²y -xyz - zx² + yx² + y³ + yz² -xy² -y²z -xyz + zx² + zy² + z³ - xyz - yz² - z²x ]

[ All square terms will cancel now. ]

→ [ x³ + y³ + z³ - 3xyz] = LHS .

✪✪ Hence Proved ✪✪

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