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) = p^{2} + q^{2}

Then,

Volume of cube = (p^{2} + q^{2})^{3}

Recall the algebraic identity,

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

Put a = p^{2} and b = q^{2}. We get

(p^{2} + q^{2})^{3} = (p^{2})^{3} + (q^{2})^{3} + 3(p^{2})^{2}(q^{2}) + 3(p^{2}) (q^{2})^{2}

⇒ (p^{2} + q^{2})^{3} = p^{6} + q^{6} + 3p^{4}q^{2} + 3p^{2}q^{4}

⇒ (p^{2} + q^{2})^{3}= p^{6} + q^{6} + 3p^{2}q^{2} (p^{2} + q^{2})

= p^{6} + q^{6} + 3p^{4}q^{2} + 3p^{2}q^{4}

Thus, volume of cube is **p ^{6} + q^{6} + 3p^{4}q^{2} + 3p^{2}q^{4}**cubic unit.

**(ii).** Given is,

Then,

Recall the algebraic identity,

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

Put and . We get

Thus, volume of cube is cubic unit.

**(iii).** Given is,

Length of cube (unit) = x^{2}y – z^{2}

Then,

Volume of cube = (x^{2}y – z^{2})^{3}

Recall the algebraic identity,

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

Put a = x^{2}y and b = z^{2}. We get

(x^{2}y – z^{2})^{3} = (x^{2}y)^{3} – (z^{2})^{3} – 3(x^{2}y)^{2}(z^{2}) + 3(x^{2}y)(z^{2})^{2}

= x^{6}y^{3} – z^{6} – 3x^{4}y^{2}z^{2} + 3x^{2}yz^{4}

= x^{6}y^{3} – z^{6} – 3x^{2}yz^{2} (x^{2}y – z^{2})

= x^{6}y^{3} – z^{6} – 3x^{4}y^{2}z^{2} + 3x^{2} y z^{4}

Thus, volume of cube is **x ^{6}y^{3} – z^{6} – 3x^{4}y^{2}z^{2} + 3x^{2} y z^{4}** 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} = a^{3} + b^{3} + c^{3} + 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 + b^{3} – 8c^{3} + 3(1 + b) (b – 2c)(1 – 2c)

Thus, volume of cube is **[1 + b ^{3} – 8c^{3} + 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 = a^{3} + b^{3} + [3 × (a + b) × a × b]

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

Or, Volume = (a + b)^{3} …(a)

[∵, by algebraic identity, we know that

(a + b)^{3} = a^{3} + b^{3} + 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 (4m^{2} – 9n^{2})

We need to find length of cube.

Volume of cube can be written as,

[∵, By identity, (a + b) (a – b) = a^{2} – b^{2} 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 = a^{3} + b^{3} + 3(a + b) ab

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

Or, Volume = (a + b)^{3} …(a)

[∵, by algebraic identity, we know that

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

We need to find the length of cube.

Volume of cube can be written as,

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

[∵, 6b = 3 × 2b]

⇒ Volume = (a + b)^{3} – (a – b)^{3} – [3 × 2b × (a + b)(a – b)]

[∵, By identity, (a^{2} – b^{2}) = (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 = x^{3} – y^{3} – 3(x – y)xy

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

Or, Volume = (x – y)^{3} …(A)

[∵, by algebraic identity, we know that

(x – y)^{3} = x^{3} – y^{3} – 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} = a^{3} + b^{3} + c^{3} + 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} = 8x^{3} – 27y^{3} – 64z^{3} + 3(2x – 3y – 4z)(-6xy + 12yz – 8zx) – 72xyz

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

**(ix).** Given is,

Volume of cube (cubic unit) = x^{6} – 15x^{4} + 75x^{2} – 125

We need to find length of cube.

Volume of cube can be written as,

[∵, x^{6} = (x^{2})^{3}, 15 = 3 × 5, 75 = 3 × 25 and 125 = (5)^{3}]

Volume = (x^{2})^{3} – (3 × 5x^{4}) + (3 × 25x^{2}) – (5)^{3}

⇒ Volume = (x^{2})^{3} – (3 × 5 × (x^{2})^{2}) + (3 × (5)^{2} × x^{2}) – (5)^{3}

[∵, x^{4} = (x^{2})^{2} and 25 = (5)^{2}]

⇒ Volume = (x^{2})^{3} + (3 × -5 × (x^{2})^{2}) + (3 × (-5)^{2} × x^{2}) + (-5)^{3} …(A)

[∵, (5)^{2} = (-5)^{2} and -(5)^{3} = (-5)^{3}]

Put x^{2} = a and (-5) = b in equation (A). We get

Volume = a^{3} + (3 × b × a^{2}) + (3 × b^{2} × a) + b^{3}

Or, Volume = a^{3} + 3ba^{2} + 3b^{2}a + b^{3}

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

[∵, by algebraic identity,

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

Put a = x^{2} and b = (-5). We get

Volume = (x^{2} – 5)^{3}

We know, volume of cube is given by

Volume = (length)^{3}

Or (length)^{3} = Volume

⇒ length = (Volume)^{1/3}

⇒ length = x^{2} – 5

Thus, length of cube is **(x ^{2} – 5)** unit.

**(x).** Given is,

Volume of cube (cubic unit) = 1000 + 30x(10 + x) + x^{3}

We need to find length of cube.

Volume of cube can be written as,

Volume = (10)^{3} + 30x(10 + x) + x^{3}

[∵, 1000 = (10)^{3}]

⇒ Volume = (10)^{3} + [3 × 10x × (10 + x)] + x^{3} …(A)

[∵, 30 = 3 × 10]

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

Volume = a^{3} + [3 × ab × (a + b)] + b^{3}

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

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

[∵, by algebraic identity,

(a + b)^{3} = a^{3} + b^{3} + 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.

Rate this question :

<img src="https:/West Bengal - Mathematics

Let us write in tWest Bengal - Mathematics

Let us simplify uWest Bengal - Mathematics

Let us solve the West Bengal - Mathematics

Let’s find the vaWest Bengal - Mathematics

Let’s find the vaWest Bengal - Mathematics

Let’s resolve intWest Bengal - Mathematics

Let’s find the vaWest Bengal - Mathematics