Q. 14.3( 4 Votes )

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Answer :

Given is, lengths of a side of cubes.

We need to find volumes of these cubes.


Let us recall the formula of volume of cube.


Volume of cube = length × length × length


Volume of cube = (length)3


(i). Given is,


Length of cube (unit) = p2 + q2


Then,


Volume of cube = (p2 + q2)3


Recall the algebraic identity,


(a + b)3 = a3 + b3 + 3a2b + 3ab2


Put a = p2 and b = q2. We get


(p2 + q2)3 = (p2)3 + (q2)3 + 3(p2)2(q2) + 3(p2) (q2)2


(p2 + q2)3 = p6 + q6 + 3p4q2 + 3p2q4


(p2 + q2)3= p6 + q6 + 3p2q2 (p2 + q2)


= p6 + q6 + 3p4q2 + 3p2q4


Thus, volume of cube is p6 + q6 + 3p4q2 + 3p2q4cubic unit.


(ii). Given is,



Then,



Recall the algebraic identity,


(a + b)3 = a3 + b3 + 3a2b + 3ab2


Put and . We get






Thus, volume of cube is cubic unit.


(iii). Given is,


Length of cube (unit) = x2y – z2


Then,


Volume of cube = (x2y – z2)3


Recall the algebraic identity,


(a – b)3 = a3 – b3 – 3a2b + 3ab2


Put a = x2y and b = z2. We get


(x2y – z2)3 = (x2y)3 – (z2)3 – 3(x2y)2(z2) + 3(x2y)(z2)2


= x6y3 – z6 – 3x4y2z2 + 3x2yz4


= x6y3 – z6 – 3x2yz2 (x2y – z2)


= x6y3 – z6 – 3x4y2z2 + 3x2 y z4


Thus, volume of cube is x6y3 – z6 – 3x4y2z2 + 3x2 y z4 cubic unit.


(iv). Given is,


Length of cube (unit) = 1 + b – 2c


Then,


Volume of cube = (1 + b – 2c)3


Recall the algebraic identity,


(a + b + c)3 = a3 + b3 + c3 + 3(a + b) (b + c)(c + a)


Put a = 1, b = b and c = -2c. We get


(1 + b – 2c)3 = (1)3 + (b)3 + (-2c)3 + 3(1 + b) (b – 2c)(-2c + 1)


(1 + b – 2c)3 = 1 + b3 – 8c3 + 3(1 + b) (b – 2c)(1 – 2c)


Thus, volume of cube is [1 + b3 – 8c3 + 3(1 + b)(b – 2c)(1 – 2c)] cubic unit.


(v). Given is,


Volume of cube (cubic unit) = (2.89)3 + (2.11)3 + 15 × 2.89 × 2.11


We need to find length of cube.


Volume of cube can be written as,


Volume = (2.89)3 + (2.11)3 + [3 × 5 × 2.89 × 2.11]


[, 15 = 3 × 5]


Volume = (2.89)3 + (2.11)3 + [3 × (2.89 + 2.11) × 2.89 × 2.11]


[, 5 = 2.89 + 2.11]


Put 2.89 = a and 2.11 = b. We get


Volume = a3 + b3 + [3 × (a + b) × a × b]


Or, Volume = a3 + b3 + 3ab (a + b)


Or, Volume = (a + b)3 …(a)


[, by algebraic identity, we know that


(a + b)3 = a3 + b3 + 3ab (a + b)]


Again,


Put a = 2.89 and b = 2.11. We get


Volume = (2.89 + 2.11)3


Volume = (5)3


We know, volume of cube is given by


Volume = (length)3


Or (length)3 = Volume


length = (Volume)1/3




length = 5


Thus, length of cube is 5 unit.


(vi). Given is,


Volume of cube (cubic unit) = (2m + 3n)3 + (2m – 3n)3 + 12m (4m2 – 9n2)


We need to find length of cube.


Volume of cube can be written as,


[, By identity, (a + b) (a – b) = a2 – b2 in (2m + 3n) (2m – 3n)]


Volume = (2m + 3n)3 + (2m – 3n)3 + [3 × 4m × (2m + 3n) (2m – 3n)]


[, 4m = 2m + 2m]


= (2m + 3n)3 + (2m – 3n)3 + [3 × (2m + 2m) × (2m + 3n) (2m – 3n)]


[, 3n – 3n = 0 and it won’t affect the equation]


Volume = (2m + 3n)3 + (2m – 3n)3 + [3 × (2m + 2m + 3n – 3n) × (2m + 3n) (2m – 3n)]


[, (2m + 2m + 3n – 3n) = (2m + 3n) + (2m – 3n); we have just rearranged]


Volume = (2m + 3n)3 + (2m – 3n)3 + [3 × ((2m + 3n) + (2m – 3n)) × (2m + 3n)(2m – 3n)]
Volume = (2m + 3n)3 + (2m – 3n)3 + 3[(2m + 3n) + (2m – 3n)][(2m + 3n)(2m – 3n)]


Put (2m + 3n) = a and (2m – 3n) = b. We get


Volume = a3 + b3 + 3(a + b) ab


Or, Volume = a3 + b3 + 3ab (a + b)


Or, Volume = (a + b)3 …(a)


[, by algebraic identity, we know that


(a + b)3 = a3 + b3 + 3ab (a + b)]


Again,


Put a = (2m + 3n) and b = (2m – 3n) in equation (a). We get


Volume = ((2m + 3n) + (2m – 3n))3


Volume = (2m + 3n + 2m – 3n)3


Volume = (4m)3


We know, volume of cube is given by


Volume = (length)3


Or (length)3 = Volume


length = (Volume)1/3




length = 4m


Thus, length of cube is 4m unit.


(vii). Given is,


Volume of cube (cubic unit) = (a + b)3 – (a – b)3 – 6b(a2 – b2)


We need to find the length of cube.


Volume of cube can be written as,


Volume = (a + b)3 – (a – b)3 – [3 × 2b × (a2 – b2)]


[, 6b = 3 × 2b]


Volume = (a + b)3 – (a – b)3 – [3 × 2b × (a + b)(a – b)]


[, By identity, (a2 – b2) = (a + b)(a – b)]


Volume = (a + b)3 – (a – b)3 – [3 × (b + b) × (a + b)(a – b)]


[, 2b = b + b]


Volume = (a + b)3 – (a – b)3 – [3 × (b + b + a – a) × (a + b)(a – b)]


[, a – a = 0 and it won’t affect the equation]


Volume = (a + b)3 – (a – b)3 – [3 × ((a + b) + (-a + b)) × (a + b)(a – b)]


[, (b + b + a – a) = ((a + b) + (-a + b)); we have just rearranged]


Volume = (a + b)3 – (a – b)3 – [3 × ((a + b) – (a – b)) × (a + b)(a – b)]


[, (-a + b) = -(a – b)]


Volume = (a + b)3 – (a – b)3 – 3((a + b) – (a – b))(a + b)(a – b)


Put (a + b) = x and (a – b) = y. We get


Volume = x3 – y3 – 3(x – y)xy


Or, Volume = x3 – y3 – 3xy(x – y)


Or, Volume = (x – y)3 …(A)


[, by algebraic identity, we know that


(x – y)3 = x3 – y3 – 3xy(x – y)]


Again,


Put x = (a + b) and y = (a – b) in equation (A). We get


Volume = ((a + b) – (a – b))3


Volume = (a + b – a + b)3


Volume = (2b)3


We know, volume of cube is given by


Volume = (length)3


Or (length)3 = Volume


length = (Volume)1/3




length = 2b


Thus, length of cube is 2b unit.


(viii). Given is,


Length of cube = 2x – 3y – 4z


Then,


Volume of cube = (2x – 3y – 4z)3


Recall the algebraic identity,


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


Put a = 2x, b = -3y and c = -4z. We get


(2x – 3y – 4z)3 = (2x)3 + (-3y)3 + (-4z)3 + 3(2x – 3y – 4z)((2x)(-3y) + (-3y)(-4z) + (-4z)(2x)) – 3(2x)(-3y)(-4z)


(2x – 3y – 4z)3 = 8x3 – 27y3 – 64z3 + 3(2x – 3y – 4z)(-6xy + 12yz – 8zx) – 72xyz


Thus, volume of cube is [8x3 – 27y3 – 64z3 + 3(2x – 3y – 4z)(-6xy + 12yz – 8zx) – 72xyz] cubic unit.


(ix). Given is,


Volume of cube (cubic unit) = x6 – 15x4 + 75x2 – 125


We need to find length of cube.


Volume of cube can be written as,


[, x6 = (x2)3, 15 = 3 × 5, 75 = 3 × 25 and 125 = (5)3]


Volume = (x2)3 – (3 × 5x4) + (3 × 25x2) – (5)3


Volume = (x2)3 – (3 × 5 × (x2)2) + (3 × (5)2 × x2) – (5)3


[, x4 = (x2)2 and 25 = (5)2]


Volume = (x2)3 + (3 × -5 × (x2)2) + (3 × (-5)2 × x2) + (-5)3 …(A)


[, (5)2 = (-5)2 and -(5)3 = (-5)3]


Put x2 = a and (-5) = b in equation (A). We get


Volume = a3 + (3 × b × a2) + (3 × b2 × a) + b3


Or, Volume = a3 + 3ba2 + 3b2a + b3


Or, Volume = (a + b)3


[, by algebraic identity,


(a + b)3 = a3 + b3 + 3a2b + 3ab2]


Put a = x2 and b = (-5). We get


Volume = (x2 – 5)3


We know, volume of cube is given by


Volume = (length)3


Or (length)3 = Volume


length = (Volume)1/3




length = x2 – 5


Thus, length of cube is (x2 – 5) unit.


(x). Given is,


Volume of cube (cubic unit) = 1000 + 30x(10 + x) + x3


We need to find length of cube.


Volume of cube can be written as,


Volume = (10)3 + 30x(10 + x) + x3


[, 1000 = (10)3]


Volume = (10)3 + [3 × 10x × (10 + x)] + x3 …(A)


[, 30 = 3 × 10]


Put 10 = a and x = b in equation (A). We get


Volume = a3 + [3 × ab × (a + b)] + b3


Or, Volume = a3 + b3 + 3ab(a + b)


Or, Volume = (a + b)3


[, by algebraic identity,


(a + b)3 = a3 + b3 + 3ab(a + b)]


Put a = 10 and b = x. We get


Volume = (10 + x)3


We know, volume of cube is given by


Volume = (length)3


Or (length)3 = Volume


length = (Volume)1/3




length = 10 + x


Thus, length of cube is (10 + x) unit.


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