Question

<img src="https:/

Goprep Admin · Accepted 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 + q2Then,Volume of cube = (p2 + q2)3Recall the algebraic identity,(a + b)3 = a3 + b3 + 3a2b + 3ab2Put 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 + 3p2q4Thus, volume of cube is p6 + q6 + 3p4q2 + 3p2q4cubic unit.(ii). Given is,Then,Recall the algebraic identity,(a + b)3 = a3 + b3 + 3a2b + 3ab2Put  and . We getThus, volume of cube is cubic unit.(iii). Given is,Length of cube (unit) = x2y – z2Then,Volume of cube = (x2y – z2)3Recall the algebraic identity,(a – b)3 = a3 – b3 – 3a2b + 3ab2Put 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 z4Thus, volume of cube is x6y3 – z6 – 3x4y2z2 + 3x2 y z4 cubic unit.(iv). Given is,Length of cube (unit) = 1 + b – 2cThen,Volume of cube = (1 + b – 2c)3Recall 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 × 2.11We 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 getVolume = 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 getVolume = (2.89 + 2.11)3⇒ Volume = (5)3We know, volume of cube is given byVolume = (length)3Or (length)3 = Volume⇒ length = (Volume)1/3⇒ length = 5Thus, 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 getVolume = a3 + b3 + 3(a + b) abOr, 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 getVolume = ((2m + 3n) + (2m – 3n))3⇒ Volume = (2m + 3n + 2m – 3n)3⇒ Volume = (4m)3We know, volume of cube is given byVolume = (length)3Or (length)3 = Volume⇒ length = (Volume)1/3⇒ length = 4mThus, 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 getVolume = x3 – y3 – 3(x – y)xyOr, 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 getVolume = ((a + b) – (a – b))3⇒ Volume = (a + b – a + b)3⇒ Volume = (2b)3We know, volume of cube is given byVolume = (length)3Or (length)3 = Volume⇒ length = (Volume)1/3⇒ length = 2bThus, length of cube is 2b unit.(viii). Given is,Length of cube = 2x – 3y – 4zThen,Volume of cube = (2x – 3y – 4z)3Recall the algebraic identity,(a + b + c)3 = a3 + b3 + c3 + 3(a + b + c)(ab + bc + ca) – 3abcPut 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) – 72xyzThus, 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 – 125We 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 getVolume = a3 + (3 × b × a2) + (3 × b2 × a) + b3Or, Volume = a3 + 3ba2 + 3b2a + b3Or, Volume = (a + b)3[∵, by algebraic identity,(a + b)3 = a3 + b3 + 3a2b + 3ab2]Put a = x2 and b = (-5). We getVolume = (x2 – 5)3We know, volume of cube is given byVolume = (length)3Or (length)3 = Volume⇒ length = (Volume)1/3⇒ length = x2 – 5Thus, length of cube is (x2 – 5) unit.(x). Given is,Volume of cube (cubic unit) = 1000 + 30x(10 + x) + x3We 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 getVolume = a3 + [3 × ab × (a + b)] + b3Or, 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 getVolume = (10 + x)3We know, volume of cube is given byVolume = (length)3Or (length)3 = Volume⇒ length = (Volume)1/3⇒ length = 10 + xThus, length of cube is (10 + x) unit.