Let us solve the questions given below with the help of identities from I to IV.

a. Let’s write the value of if

b. Let’s try to prove if

c. Let’s write the value of by calculation if and

d. Let’s try to write the value of if

e. Let’s write the value of by calculation if

f. Let’s write the value of if

g. Let’s try to prove if

h. Let’s try to prove if

i. Let’s write the value of if

j. Let’s write the value of by calculation if

k. Let’s write the value of by calculation if and

Let us solve.

(a). Given that, x – y = 2.

We need to find the value of x³ – y³ – 6xy.

By algebraic identity,

(a – b)³ = a³ – b³ – 3a²b + 3ab²

Or, (a – b)³ = a³ – b³ – 3ab(a – b)

Or, a³ – b³ = (a – b)³ + 3ab(a – b)

Put a = x and b = y. We get

x³ – y³ = (x – y)³ + 3xy(x – y) …(i)

Take x³ – y³ – 6xy.

x³ – y³ – 6xy = (x³ – y³) – 6xy

⇒ x³ – y³ – 6xy = ((x – y)³ + 3xy(x – y)) – 6xy [from (i)]

⇒ x³ – y³ – 6xy = ((2)³ + 3xy(2)) – 6xy [∵, Given that, (x – y) = 2]

⇒ x³ – y³ – 6xy = (8 + 6xy) – 6xy [∵, 2³ = 8]

⇒ x³ – y³ – 6xy = 8 + 6xy – 6xy

⇒ x³ – y³ – 6xy = 8 + 0 [∵, 6xy – 6xy = 0]

⇒ x³ – y³ – 6xy = 8

Thus, x³ – y³ – 6xy = 8.

(b). Given:

To prove:

Proof: We know the algebraic identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

Or, (a + b)³ = a³ + b³ + 3ab(a + b)

Or, a³ + b³ = (a + b)³ – 3ab(a + b) …(i)

Take a³ + b³ – ab.

a³ + b³ – ab = (a³ + b³) – ab

⇒ a³ + b³ – ab = ((a + b)³ – 3ab(a + b)) – ab

[∵, we know that, ]

Thus, .

Hence proved.

(c). Given that, x + y = 2 and

Let us solve this further.

We have,

⇒ x + y = 2xy

⇒ 2xy = x + y

⇒ 2xy = 2 [∵, we know that, (x + y) = 2]

⇒ xy = 1

Now, we have

x + y = 2 …(i)

xy = 1 …(ii)

We need to find the value of x³ + y³.

By algebraic identity,

(x + y)³ = x³ + y³ + 3x²y + 3xy²

Or, (x + y)³ = x³ + y³ + 3xy(x + y)

Or, x³ + y³ = (x + y)³ – 3xy(x + y)

Substituting value of (x + y) and xy from equation (i) and (ii) respectively, we get

x³ + y³ = (2)³ – 3(1)(2)

⇒ x³ + y³ = 8 – 6

⇒ x³ + y³ = 2

Thus, x³ + y³ = 2.

(d). Given that,

…(i)

We need to find the value of

Further simplifying (i), we get

x² – 1 = 2x

Now, take cube on both sides. We get

(x² – 1)³ = (2x)³

[∵, we have the identity,

(a – b)³ = a³ – b³ – 3a²b + 3ab²]

⇒ (x²)³ – 1³ – 3(x²)² + 3x² = (2x)³

⇒ x⁶ – 1 – 3x⁴ + 3x² = 8x³

Dividing both sides by x³,

Thus, .

(e). Given that,

We need to find the value of .

Let us take .

Take cube on both sides of this equation. We get

[∵, from the identity, (a + b)³ = a³ + b³ + 3a²b + 3ab²]

Thus, .

(f). Given that,

x = y + z

We need to find the value of x³ – y³ – z³ – 3xyz.

Take x = y + z.

We can write as,

x – y – z = 0 …(i)

Now, take cube on both sides of the above equation. We get,

(x – y – z)³ = 0

Since, by the identity

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

Rearranging it,

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

⇒ x³ – y³ – z³ – 3xyz = (0)³ – 3(0)(-xy + yz – zx) [from equation (i)]

⇒ x³ – y³ – z³ – 3xyz = 0 – 0

⇒ x³ – y³ – z³ – 3xyz = 0

Thus, x³ – y³ – z³ – 3xyz = 0.

(g). Given: xy(x + y) = m

To prove:

Proof: Take xy(x + y) = m.

Rearranging it,

Taking cube on both sides, we get

By identity, we have

(x + y)³ = x³ + y³ + 3x²y + 3xy²

So,

[∵, given that, xy(x + y) = m]

Hence, proved.

(h). Given:

To prove:

Proof: We have,

Putting and .

⇒ p + q = 4pq …(i)

We have,

Multiplying p and q, we get

…(ii)

Now, take .

[∵, (p + q)³ = p³ + q³ + 3p²q + 3pq²

Or, p³ + q³ = (p + q)³ – 3p²q – 3pq²

Or, p³ + q³ = (p + q)³ – 3pq(p + q)]

…(iii)

Substituting the value of pq from equation (ii) in equation (iii), we get

Hence, proved.

(i). Given that,

We need to find the value of .

We have,

…(i)

Taking cube on both sides of the equation, we get

Now, adding 2 on both sides of this equation, we get

Thus, .

(j). Given that,

a³ + b³ + c³ = 3abc

where a ≠ b ≠ c.

We need to find the value of a + b + c.

We have,

a³ + b³ + c³ = 3abc

⇒ a³ + b³ + c³ – 3abc = 0

We know that,

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

⇒ 0 = (a + b + c)(a² + b² + c² – ab – bc – ca)

⇒ (a + b + c)(a² + b² + c² – ab – bc – ca) = 0

This means,

Either (a + b + c) = 0

Or (a² + b² + c² – ab – bc – ca) = 0

If a² + b² + c² – ab – bc – ca = 0

Multiplying by 2 on both sides, we get

2(a² + b² + c² – ab – bc – ca) = 2 × 0

⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

⇒ a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ca = 0

⇒ (a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² + 2ca) = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0

[∵, by identity, we know (x – y)² = x² + y² – 2xy]

This means,

(a – b)² = 0 ⇒ a = b

(b – c)² = 0 ⇒ b = c

(c – a)² = 0 ⇒ c = a

But, a ≠ b ≠ c.

⇒ (a² + b² + c² – ab – bc – ca) ≠ 0

This clearly means,

a + b + c = 0

Thus, a + b + c = 0.

(k). Given that,

m + n = 5 …(i)

mn = 6 …(ii)

We need to find the value of (m² + n²)(m³ + n³).

By identity, we know

(m + n)³ = m³ + n³ + 3mn(m + n)

Or, m³ + n³ = (m + n)³ – 3mn(m + n) …(iii)

Also, by identity,

(m + n)² = m² + n² + 2mn

Or, m² + n² = (m + n)² – 2mn …(iv)

Take (m² + n²)(m³ + n³).

So,

(m² + n²)(m³ + n³) = ((m + n)² – 2mn)((m + n)³ – 3mn(m + n))

Substituting values of (m + n) and mn from equation (i) and (ii) respectively in the above equation, we get

(m² + n²)(m³ + n³) = ((5)² – 2(6))((5)³ – 3(6)(5))

⇒ (m² + n²)(m³ + n³) = (25 – 12)(125 – 90)

⇒ (m² + n²)(m³ + n³) = 13 × 35

⇒ (m² + n²)(m³ + n³) = 455

Thus, (m² + n²)(m³ + n³) = 455.