Three brackets are done as two brackets and then one. Take the first pair:

\((x + 3)(x + 2) = x^2 + 5x + 6\)

Now multiply every term of that quadratic by every term of \((x + 1)\):

\(x^2 \times x = x^3\)

\(x^2 \times 1 = x^2\)

\(5x \times x = 5x^2\)

\(5x \times 1 = 5x\)

\(6 \times x = 6x\)

\(6 \times 1 = 6\)

Collect the two \(x^2\) terms and the two \(x\) terms:

\(x^2 + 5x^2 = 6x^2\)

\(5x + 6x = 11x\)

\(x^3 + 6x^2 + 11x + 6\)

A check worth doing every time: the number on the end is always the three constants multiplied together, \(3 \times 2 \times 1 = 6\).

Deal with the squared bracket first. Squaring a bracket means multiplying it by itself, so

\((x - 5)^2 = x^2 - 10x + 25\)

The question is now a quadratic times a linear bracket. Multiply every term of one by every term of the other:

\(x^2 \times x = x^3\)

\(x^2 \times 6 = 6x^2\)

\(-10x \times x = -10x^2\)

\(-10x \times 6 = -60x\)

\(25 \times x = 25x\)

\(25 \times 6 = 150\)

Collect:

\(6x^2 - 10x^2 = -4x^2\)

\(-60x + 25x = -35x\)

\(x^3 - 4x^2 - 35x + 150\)

The volume of a cuboid is length \(\times\) width \(\times\) height, so the volume is \((x + 2)(x + 5)(x + 1)\) cm³.

Multiply the first two brackets:

\((x + 2)(x + 5) = x^2 + 7x + 10\)

Then multiply that by \((x + 1)\):

\(x^2 \times x = x^3\), \(\quad x^2 \times 1 = x^2\)

\(7x \times x = 7x^2\), \(\quad 7x \times 1 = 7x\)

\(10 \times x = 10x\), \(\quad 10 \times 1 = 10\)

Collecting the \(x^2\) terms and the \(x\) terms:

\(x^3 + 8x^2 + 17x + 10\)

Put \(x = 2\) into the expression from part (a):

\(8 + 32 + 34 + 10 = 84\)

It is worth checking that against the three dimensions themselves, which are \(4\) cm, \(7\) cm and \(3\) cm when \(x = 2\):

\(4 \times 7 \times 3 = 84\)

The two agree, which is what "expand and simplify" is supposed to guarantee — the expression and the product are the same number for every value of \(x\).

The constant term of the expansion comes from multiplying the three constants together — every other term has at least one \(x\) in it.

\(6 \times (-1) \times k = -6\)

\(-6k = -6\)

\(k = -6 \div -6 = 1\)

Multiply every term of the second bracket by \(x\), and then every term of it by \(1\). Two rows, three terms each.

\(x(x^2 +x - 7) = x^3 +x^2 - 7x\)

\(1(x^2 +x - 7) = x^2 +x - 7\)

Now add the two rows and collect. There is one \(x^3\), the \(x^2\) terms give \(1 + 1 = 2\), the \(x\) terms give \(-7 + 1 = -6\), and the number on the end comes only from the second row.

\(x^3 + 2x^2 - 6x - 7\)

The check worth doing: the constant term of an expansion is always the product of the constants, \(1 \times (-7) = -7\), and the \(x^3\) term always comes from multiplying the two leading terms.

A bracket cubed is that bracket three times over, so square it first and then multiply by the bracket once more.

\((w + 1)^2 = w^2 + 2w + 1\)

Now multiply that by \((w + 1)\), one row for the \(w\) and one for the \(1\):

\(w(w^2 + 2w + 1) = w^3 + 2w^2 + w\)

\(1(w^2 + 2w + 1) = w^2 + 2w + 1\)

Add the two rows. The \(w^2\) terms give \(2 + 1 = 3\), and the \(w\) terms give \(1 + 2 = 3\):

\(w^3 + 3w^2 + 3w + 1\)

The pattern is worth remembering: \((x + a)^3 = x^3 + 3ax^2 + 3a^2x + a^3\), and the \(3a^2\) term is positive whether \(a\) is positive or negative, because it is a square.

Deal with the two brackets first and leave the \(2x\) outside until the end — it is one multiplication either way, and doing it last keeps the numbers small.

\((x + 7)(x - 5) = x^2 + 2x - 35\)

Now multiply every term of that by \(2x\). Multiplying by \(x\) raises each power by one:

\(2x \times x^2 = 2x^3\)

\(2x \times 2x = 4x^2\)

\(2x \times (-35) = -70x\)

\(2x^3 + 4x^2 - 70x\)

There is no number on the end. Every term had an \(x\) put into it, so every term still has one — which is also why the whole thing factorises back to \(2x(\dots)\).

An \(x^2\) term is made by taking the \(x\) from two of the brackets and the number from the third, and there are three ways to choose which bracket gives the number. So \(p\) is the three numbers added:

\(p = -3 + 2 - 1 = -2\)

An \(x\) term is made by taking the \(x\) from one bracket and the number from the other two, so \(q\) is every pair of numbers multiplied together and then added:

\((-3) \times 2 = -6\)

\((-3) \times (-1) = 3\)

\(2 \times (-1) = -2\)

\(q = -6 + 3 - 2 = -5\)

The number on the end can only come from taking the number out of all three brackets, so \(r\) is the three of them multiplied:

\(r = (-3) \times 2 \times (-1) = 6\)

Multiply the two ordinary brackets first — leaving the awkward one until last means the coefficient only has to be handled once.

\((x - 3)(x + 2) = x^2 -x - 6\)

Now multiply that quadratic by \((2x + 3)\). Each of the two terms in the small bracket hits all three terms in the quadratic:

\(2x \times x^2 = 2x^3\)

\(2x \times (-x) = -2x^2\)

\(2x \times (-6) = -12x\)

\(3 \times x^2 = 3x^2\)

\(3 \times (-x) = -3x\)

\(3 \times (-6) = -18\)

Collect the \(x^2\) terms and the \(x\) terms:

\(-2x^2 + 3x^2 = x^2\)

\(-12x - 3x = -15x\)

\(2x^3 +x^2 - 15x - 18\)

Find the three dimensions of the box first.

The height is the depth of the fold, which is \(x\) cm.

The long side of the card loses a square at each end, so the base measures \(17 - 2x\) cm one way. The short side loses one at each end too, so it measures \(11 - 2x\) cm the other way.

\(\text{Volume} = x(17 - 2x)(11 - 2x)\)

Multiply the two brackets together first:

\((17 - 2x)(11 - 2x) = 187 - 34x - 22x + 4x^2 = 4x^2 - 56x + 187\)

Now multiply every term by \(x\):

\(4x^3 - 56x^2 + 187x\)

A quick check on the shape of the answer: every term has an \(x\) in it, because the height is \(x\) and a box of zero height has no volume.