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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