Key idea. A question asking whether something can happen is settled by one working example.

1. Try the three positive integers 1, 2 and 3. Their sum is \(1+2+3=6\).

2. Their product is \(1\cdot2\cdot3=6\), so it is equal to their sum.

3. All three numbers are positive integers, as required. We have checked every condition of the example.

Remember. One example proves that something is possible. We do not need to list every possible example.

Key idea. Read “Every champion does well” in the direction it is stated.

1. The mother’s statement tells us something about champions: they all belong to the group of pupils who do well at school.

2. It does not say that champions are the only pupils in that group. Someone may do well without being a champion.

3. Imagine Kolya is a pupil who does well and is not a champion. This agrees with his mother’s statement but makes his own conclusion false. Therefore the conclusion does not follow.

Remember. “Every A is B” does not automatically mean “Every B is A”.

Key idea. Look for two divisors that share a factor.

1. Choose \(n=12\), \(a=4\) and \(b=6\). Both divisibility conditions hold: \(12=4\cdot3=6\cdot2\).

2. However, \(ab=4\cdot6=24\), and 12 is not divisible by 24.

3. This single example shows that the situation asked about can happen.

Remember. Checking the two divisors separately does not always justify multiplying them.

Key idea. Begin with the resident whose existence is guaranteed.

1. We know that someone in the town has beautiful handwriting. Choose such a person.

2. This person cannot be a poet, because no poet has beautiful handwriting.

3. The person cannot be a boxer either: every boxer is a poet, and we have already ruled out being a poet.

4. So there is a resident who is neither a boxer nor a poet. That contradicts the visitor’s claim about every resident.

Remember. To disprove a claim about everybody, find one person who cannot fit it.

Key idea. Assume there is a hippo and follow the consequences.

1. The zoo is known to have a giraffe. Suppose it also has a hippo.

2. The rule about hippos and giraffes would then force it to have a rhino.

3. But a zoo with both a hippo and a rhino cannot have a giraffe. This contradicts the giraffe that is already there. So our assumption was impossible: there is no hippo.

4. Every zoo must have at least a rhino or a hippo. Since this zoo has no hippo, it must have a rhino.

5. A giraffe and a rhino, with no hippo, cause no conflict with any of the three rules.

Remember. A contradiction rules out the assumption that led to it.

Key idea. Use a difference of squares on the two whole brackets.

1. The identity is \(U^2-V^2=(U-V)(U+V)\). Here take \(U=2a+b\) and \(V=a-2b\).

2. Find the difference carefully:

   \[U-V=(2a+b)-(a-2b)=2a+b-a+2b=a+3b.\]

3. Find the sum:

   \[U+V=(2a+b)+(a-2b)=3a-b.\]

4. Multiply the two expressions in factorised form. This gives \((a+3b)(3a-b)\).

Remember. A complicated-looking bracket can be treated as a single quantity. Be careful when subtracting a bracket.

Key idea. Express the required quantities using the known sum and product.

1. Expand the square of the sum:

   \[(x+y)^2=x^2+2xy+y^2.\]

   So \(x^2+y^2=7^2-2\cdot10=29\).

2. For the cubes, expand \((x+y)(x^2-xy+y^2)\). The mixed terms cancel, leaving \(x^3+y^3\).

3. Therefore

   \[x^3+y^3=(x+y)(x^2+y^2-xy)=7(29-10)=133.\]

Remember. Find an identity that uses what you know; solving for the individual numbers may be unnecessary.

Key idea. The expression is almost the expansion of a cube.

1. Expand:

   \[(x+y)^3=x^3+3x^2y+3xy^2+y^3.\]

2. Factor the two middle terms: \(3x^2y+3xy^2=3xy(x+y)\). Since \(x+y=1\), these terms together equal \(3xy\).

3. Thus \(x^3+y^3+3xy=(x+y)^3=1^3=1\).

Remember. Before substituting individual numbers, compare an expression with a familiar identity.

Key idea. Expand both sides and use the fact that real squares are non-negative.

1. The left side expands to

   \[2a^2+2b^2+2c^2+2ab+2bc+2ac.\]

2. The right side expands to

   \[a^2+b^2+c^2+2ab+2bc+2ac.\]

3. Cancel the same mixed terms on both sides, then subtract \(a^2+b^2+c^2\). The result is \(a^2+b^2+c^2=0\).

4. Each square is at least zero. If even one were positive, their sum would be positive. Therefore all three squares, and all three numbers, must be zero.

5. Substituting \(a=b=c=0\) makes both original sides zero, so this triple really works.

Remember. A sum of real squares is zero only when every square is zero.

Key idea. Equal sums and equal sums of squares force equal products.

1. Write the common sum as \(s=a+b=c+d\). Squaring gives

   \[s^2=a^2+b^2+2ab=c^2+d^2+2cd.\]

2. The sums of squares are equal by the question. Subtract them from the equality above to obtain \(2ab=2cd\), so \(ab=cd\). Call this common product \(p\).

3. From the cube identity,

   \[a^3+b^3=(a+b)^3-3ab(a+b)=s^3-3ps.\]

4. Exactly the same calculation gives \(c^3+d^3=s^3-3ps\). Hence the two sums of cubes are equal.

Remember. Two pairs with the same sum and product have the same sum of cubes.

Key idea. Square the given sum before trying to find x itself.

1. Squaring the given equality gives

   \[\left(x+\frac1x\right)^2=9.\]

2. The middle term is \(2x\cdot(1/x)=2\), so

   \[x^2+2+\frac1{x^2}=9.\]

3. Subtract 2. Thus \(x^2+1/x^2=7\). We did not need the individual value of x.

Remember. A symmetric expression can often be found directly from another symmetric expression.

Key idea. Use the sum-of-cubes identity with reciprocal terms.

1. For any u and v, expanding \((u+v)^3\) gives

   \[u^3+v^3=(u+v)^3-3uv(u+v).\]

2. Here \(u=x\), \(v=1/x\). We know \(u+v=3\) and \(uv=1\).

3. Substitution gives

   \[x^3+\frac1{x^3}=3^3-3\cdot1\cdot3=27-9=18.\]

Remember. Knowing the sum and product of two quantities is enough to find their sum of cubes.

Key idea. Use the zero sum twice: first by squaring, then by substituting for c.

1. Square \(a+b+c=0\):

   \[a^2+b^2+c^2+2ab+2bc+2ca=0.\]

   Rearranging proves \(a^2+b^2+c^2=-2(ab+bc+ca)\).

2. For the cubes, write \(c=-(a+b)\). Then

   \[a^3+b^3+c^3=a^3+b^3-(a+b)^3.\]

3. Expand the last cube and cancel:

   \[a^3+b^3+c^3=-3a^2b-3ab^2=-3ab(a+b).\]

4. Since \(a+b=-c\), this equals \(3abc\), proving the second identity.

Remember. A relation between variables can make otherwise complicated expansions collapse.

Key idea. First find the mixed products, then square the known sum of squares.

1. By C1.3,

   \[ab+bc+ca=-\frac12(a^2+b^2+c^2)=-\frac12.\]

2. Square this relation:

   \[\begin{aligned}\frac14&=(ab+bc+ca)^2\\&=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c).\end{aligned}\]

   The final term is zero, so \(a^2b^2+b^2c^2+c^2a^2=1/4\).

3. Now square \(a^2+b^2+c^2=1\):

   \[1=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2).\]

4. The mixed part equals \(2\cdot(1/4)=1/2\). Subtracting it leaves \(a^4+b^4+c^4=1/2\).

Remember. When squaring creates mixed terms, look for a second identity that evaluates those terms.

Key idea. To prove expressions equal, factor their difference.

1. Use \(z=-(x+y)\) in the sum of squares:

   \[2=x^2+y^2+(x+y)^2=2(x^2+xy+y^2).\]

   Hence \(x^2+xy+y^2=1\).

2. Subtract two of the expressions:

   \[\begin{aligned}(x^3-x)-(y^3-y)&=(x^3-y^3)-(x-y)\\&=(x-y)(x^2+xy+y^2-1)\\&=0.\end{aligned}\]

   Thus \(x^3-x=y^3-y\).

3. The hypotheses treat x, y and z in the same way. Repeating the argument with y and z gives \(y^3-y=z^3-z\). All three expressions are therefore equal.

Remember. Equality can be easier to prove by showing that a difference is zero.