Q. 23.7( 9 Votes )

# Let us solve the

Answer :

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.

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