Exploring Algebraic Identities | Exercise 4.3

Question 4

Is this an identity?

(a+b−c)2+(a−b+c)2+(a−b−c)2=2a2+2b2+2c2(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2.

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Solution

Let's expand the left-hand side of the given equation.

Step 1 — Expand the first term

We use the identity (x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx. Here, x=ax=\mathbf{a}, y=by=\mathbf{b}, and z=−cz=\mathbf{-c}.

(a+b−c)2=a2+b2+(−c)2+2(a)(b)+2(b)(−c)+2(a)(−c)(a + b - c)^2 = a^2 + b^2 + (-c)^2 + 2(a)(b) + 2(b)(-c) + 2(a)(-c)

=a2+b2+c2+2ab−2bc−2ac= a^2 + b^2 + c^2 + 2ab - 2bc - 2ac

(a+b−c)2=a2+b2+c2+2ab−2bc−2ac\boxed{(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ac}

Diagram 1

Step 2 — Expand the second term

We apply the same identity. Here, x=ax=\mathbf{a}, y=−by=\mathbf{-b}, and z=cz=\mathbf{c}.

(a−b+c)2=a2+(−b)2+c2+2(a)(−b)+2(−b)(c)+2(a)(c)(a - b + c)^2 = a^2 + (-b)^2 + c^2 + 2(a)(-b) + 2(-b)(c) + 2(a)(c)

=a2+b2+c2−2ab−2bc+2ac= a^2 + b^2 + c^2 - 2ab - 2bc + 2ac

(a−b+c)2=a2+b2+c2−2ab−2bc+2ac\boxed{(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac}

Step 3 — Expand the third term

Again, we use the identity. Here, x=ax=\mathbf{a}, y=−by=\mathbf{-b}, and z=−cz=\mathbf{-c}.

(a−b−c)2=a2+(−b)2+(−c)2+2(a)(−b)+2(−b)(−c)+2(a)(−c)(a - b - c)^2 = a^2 + (-b)^2 + (-c)^2 + 2(a)(-b) + 2(-b)(-c) + 2(a)(-c)

=a2+b2+c2−2ab+2bc−2ac= a^2 + b^2 + c^2 - 2ab + 2bc - 2ac

(a−b−c)2=a2+b2+c2−2ab+2bc−2ac\boxed{(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ac}

Step 4 — Add all expanded terms

Let's add the results from Step 1, Step 2, and Step 3. This gives us the Left Hand Side (LHS).

LHS=(a2+b2+c2+2ab−2bc−2ac)\text{LHS} = (a^2 + b^2 + c^2 + 2ab - 2bc - 2ac) +(a2+b2+c2−2ab−2bc+2ac)+ (a^2 + b^2 + c^2 - 2ab - 2bc + 2ac) +(a2+b2+c2−2ab+2bc−2ac)+ (a^2 + b^2 + c^2 - 2ab + 2bc - 2ac)

=(a2+a2+a2)+(b2+b2+b2)+(c2+c2+c2)= (a^2 + a^2 + a^2) + (b^2 + b^2 + b^2) + (c^2 + c^2 + c^2) +(2ab−2ab−2ab)+(−2bc−2bc+2bc)+(−2ac+2ac−2ac)+ (2ab - 2ab - 2ab) + (-2bc - 2bc + 2bc) + (-2ac + 2ac - 2ac)

=3a2+3b2+3c2−2ab−2bc−2ac= 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ac

LHS=3a2+3b2+3c2−2ab−2bc−2ac\boxed{\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ac}

Step 5 — Compare LHS with RHS

The Right Hand Side (RHS) is given. RHS =2a2+2b2+2c2= \mathbf{2a^2 + 2b^2 + 2c^2}.

We compare our calculated LHS with the RHS. Our LHS is 3a2+3b2+3c2−2ab−2bc−2ac3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ac. The RHS is 2a2+2b2+2c22a^2 + 2b^2 + 2c^2. These two expressions are not equal. For example, if we choose a=1a=\mathbf{1}, b=0b=\mathbf{0}, c=0c=\mathbf{0}: LHS =3(1)2+3(0)2+3(0)2−2(1)(0)−2(0)(0)−2(1)(0)=3= 3(1)^2 + 3(0)^2 + 3(0)^2 - 2(1)(0) - 2(0)(0) - 2(1)(0) = \mathbf{3}. RHS =2(1)2+2(0)2+2(0)2=2= 2(1)^2 + 2(0)^2 + 2(0)^2 = \mathbf{2}. Since 3≠23 \neq 2, the statement is not true for all values.

Answer

(i) The expanded Left Hand Side is 3a2+3b2+3c2−2ab−2bc−2ac3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ac. (ii) The Right Hand Side is 2a2+2b2+2c22a^2 + 2b^2 + 2c^2. (iii) The given statement is not an identity.

More questions in Exercise 4.3

Q1

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i) 1172117^2

(ii) 78278^2

(iii) 1982198^2

(iv) 2142214^2

(v) 110421104^2

(vi) 112021120^2

Q2

Factor using suitable identities:

(i) 16y2−24y+916y^2 - 24y + 9

(ii) 94s2+6st+4t2\frac{9}{4}s^2 + 6st + 4t^2

(iii) m29+mk3+k24+3nk+2mn+9n2\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2

(iv) p216−2+16p2\frac{p^2}{16} - 2 + \frac{16}{p^2}

(v) 9a2+4b2+c2−12ab+6ac−4bc9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc

Q3

Expand the following using the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca:

(i) (p+3q+7r)2(p + 3q + 7r)^2

(ii) (3x−2y+4z)2(3x - 2y + 4z)^2

Q4

Is this an identity?

(a+b−c)2+(a−b+c)2+(a−b−c)2=2a2+2b2+2c2(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2.

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