Answer :

Let us solve.

**(a).** Given that, x – y = 2.

We need to find the value of x^{3} – y^{3} – 6xy.

By algebraic identity,

(a – b)^{3} = a^{3} – b^{3} – 3a^{2}b + 3ab^{2}

Or, (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

Or, a^{3} – b^{3} = (a – b)^{3} + 3ab(a – b)

Put a = x and b = y. We get

x^{3} – y^{3} = (x – y)^{3} + 3xy(x – y) …(i)

Take x^{3} – y^{3} – 6xy.

x^{3} – y^{3} – 6xy = (x^{3} – y^{3}) – 6xy

⇒ x^{3} – y^{3} – 6xy = ((x – y)^{3} + 3xy(x – y)) – 6xy [from (i)]

⇒ x^{3} – y^{3} – 6xy = ((2)^{3} + 3xy(2)) – 6xy [∵, Given that, (x – y) = 2]

⇒ x^{3} – y^{3} – 6xy = (8 + 6xy) – 6xy [∵, 2^{3} = 8]

⇒ x^{3} – y^{3} – 6xy = 8 + 6xy – 6xy

⇒ x^{3} – y^{3} – 6xy = 8 + 0 [∵, 6xy – 6xy = 0]

⇒ x^{3} – y^{3} – 6xy = 8

Thus, **x ^{3} – y^{3} – 6xy = 8**.

**(b).** Given:

To prove:

Proof: We know the algebraic identity,

(a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

Or, (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

Or, a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b) …(i)

Take a^{3} + b^{3} – ab.

a^{3} + b^{3} – ab = (a^{3} + b^{3}) – ab

⇒ a^{3} + b^{3} – 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 x^{3} + y^{3}.

By algebraic identity,

(x + y)^{3} = x^{3} + y^{3} + 3x^{2}y + 3xy^{2}

Or, (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)

Or, x^{3} + y^{3} = (x + y)^{3} – 3xy(x + y)

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

x^{3} + y^{3} = (2)^{3} – 3(1)(2)

⇒ x^{3} + y^{3} = 8 – 6

⇒ x^{3} + y^{3} = 2

Thus, **x ^{3} + y^{3} = 2**.

**(d).** Given that,

…(i)

We need to find the value of

Further simplifying (i), we get

x^{2} – 1 = 2x

Now, take cube on both sides. We get

(x^{2} – 1)^{3} = (2x)^{3}

[∵, we have the identity,

(a – b)^{3} = a^{3} – b^{3} – 3a^{2}b + 3ab^{2}]

⇒ (x^{2})^{3} – 1^{3} – 3(x^{2})^{2} + 3x^{2} = (2x)^{3}

⇒ x^{6} – 1 – 3x^{4} + 3x^{2} = 8x^{3}

Dividing both sides by x^{3},

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} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}]

Thus, .

**(f).** Given that,

x = y + z

We need to find the value of x^{3} – y^{3} – z^{3} – 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} = x^{3} – y^{3} – z^{3} + 3(x – y – z)(-xy + yz – zx) – 3xyz

Rearranging it,

x^{3} – y^{3} – z^{3} – 3xyz = (x – y – z)^{3} – 3(x – y – z)(-xy + yz – zx)

⇒ x^{3} – y^{3} – z^{3} – 3xyz = (0)^{3} – 3(0)(-xy + yz – zx) [from equation (i)]

⇒ x^{3} – y^{3} – z^{3} – 3xyz = 0 – 0

⇒ x^{3} – y^{3} – z^{3} – 3xyz = 0

Thus, **x ^{3} – y^{3} – z^{3} – 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} = x^{3} + y^{3} + 3x^{2}y + 3xy^{2}

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} = p^{3} + q^{3} + 3p^{2}q + 3pq^{2}

Or, p^{3} + q^{3} = (p + q)^{3} – 3p^{2}q – 3pq^{2}

Or, p^{3} + q^{3} = (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,

a^{3} + b^{3} + c^{3} = 3abc

where a ≠ b ≠ c.

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

We have,

a^{3} + b^{3} + c^{3} = 3abc

⇒ a^{3} + b^{3} + c^{3} – 3abc = 0

We know that,

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

⇒ 0 = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

⇒ (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca) = 0

This means,

Either (a + b + c) = 0

Or (a^{2} + b^{2} + c^{2} – ab – bc – ca) = 0

If a^{2} + b^{2} + c^{2} – ab – bc – ca = 0

Multiplying by 2 on both sides, we get

2(a^{2} + b^{2} + c^{2} – ab – bc – ca) = 2 × 0

⇒ 2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ca = 0

⇒ a^{2} + a^{2} + b^{2} + b^{2} + c^{2} + c^{2} – 2ab – 2bc – 2ca = 0

⇒ (a^{2} + b^{2} – 2ab) + (b^{2} + c^{2} – 2bc) + (c^{2} + a^{2} + 2ca) = 0

⇒ (a – b)^{2} + (b – c)^{2} + (c – a)^{2} = 0

[∵, by identity, we know (x – y)^{2} = x^{2} + y^{2} – 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.

⇒ (a^{2} + b^{2} + c^{2} – 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 (m^{2} + n^{2})(m^{3} + n^{3}).

By identity, we know

(m + n)^{3} = m^{3} + n^{3} + 3mn(m + n)

Or, m^{3} + n^{3} = (m + n)^{3} – 3mn(m + n) …(iii)

Also, by identity,

(m + n)^{2} = m^{2} + n^{2} + 2mn

Or, m^{2} + n^{2} = (m + n)^{2} – 2mn …(iv)

Take (m^{2} + n^{2})(m^{3} + n^{3}).

So,

(m^{2} + n^{2})(m^{3} + n^{3}) = ((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

(m^{2} + n^{2})(m^{3} + n^{3}) = ((5)^{2} – 2(6))((5)^{3} – 3(6)(5))

⇒ (m^{2} + n^{2})(m^{3} + n^{3}) = (25 – 12)(125 – 90)

⇒ (m^{2} + n^{2})(m^{3} + n^{3}) = 13 × 35

⇒ (m^{2} + n^{2})(m^{3} + n^{3}) = 455

Thus, **(m ^{2} + n^{2})(m^{3} + n^{3}) = 455**.

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