A Formula That Lives Inside Every Cube's Diagonal
The space diagonal of a rectangular box hides this identity at its core.
The a²+b²+c² formula is the cleanest way to compute a sum of three squares when what you know is the sum $a+b+c$ and the pairwise-product sum $ab+bc+ca$ — a situation that comes up constantly in symmetric algebra and Vieta's formulas.
The Formula
For any real numbers $a$, $b$, $c$:
$$\boxed{;a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca);}$$
Equivalently, rearranged from the expansion of $(a + b + c)^2$:
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).$$
The two forms are the same identity. Whichever is more useful depends on what you already know.
Quick facts.
Type: algebraic identity (true for all real $a$, $b$, $c$).
Reads as: "sum of three squares equals the square of the sum minus twice the sum of pairwise products."
Grade introduced: CCSS-M HSA-SSE.A.2 (seeing structure in expressions); NCERT Class 9 Chapter 2 — Polynomials Identity VIII.
Related identities: $a^2 + b^2 = (a+b)^2 - 2ab$; $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
3D Pythagorean version: for a box of sides $a, b, c$, the diagonal is $\sqrt{a^2+b^2+c^2}$.
How the Formula Is Derived — A Two-Line Proof
Expand $(a+b+c)^2$ as $(a+b+c)(a+b+c)$ and collect terms:
$$(a+b+c)^2 = a \cdot a + a \cdot b + a \cdot c + b \cdot a + b \cdot b + b \cdot c + c \cdot a + c \cdot b + c \cdot c.$$
The nine terms include three squares and six pairwise products (each pair counted twice):
$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).$$
Subtract $2(ab+bc+ca)$ from both sides:
$$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca).$$
Done. Two lines, no trickery — just the distributive property applied carefully.
Three Worked Examples, From Quick to Stretch
Quick. Find $a^2 + b^2 + c^2$ when $a = 1$, $b = 2$, $c = 3$.
Direct compute: $1 + 4 + 9 = 14$. Verify with the formula:
$a + b + c = 6$, so $(a+b+c)^2 = 36$. And $ab + bc + ca = 2 + 6 + 3 = 11$, so $2 \cdot 11 = 22$.
$$a^2 + b^2 + c^2 = 36 - 22 = 14. \checkmark$$
Final answer: $14$.
Standard (Wrong-Path-First). Given $a + b + c = 10$ and $ab + bc + ca = 27$, find $a^2 + b^2 + c^2$.
Wrong path. A student in our McKinney TX Grade 9 cohort wrote: "$(a+b+c)^2 = 100$, and $ab+bc+ca = 27$, so $a^2+b^2+c^2 = 100 - 27 = 73$." Look at the formula again — the term subtracted is $2(ab+bc+ca)$, not $(ab+bc+ca)$. Forgetting the factor of $2$ is the single most common slip on this identity. The answer $73$ is off by $27$.
Correct.
$$a^2 + b^2 + c^2 = 10^2 - 2 \cdot 27 = 100 - 54 = 46.$$
Final answer: $46$.
Stretch. A room is a rectangular box. The sum of its length, width, and height is $24$ m; the sum of the areas of its three distinct faces (each pair of opposite faces shares the same area) is $216$ m². What is the length of the space diagonal?
Let $a, b, c$ be length, width, height. Given:
$a + b + c = 24$
Sum of pairwise face areas: $2ab + 2bc + 2ca = 216$, so $ab + bc + ca = 108$.
The space diagonal $d$ satisfies $d^2 = a^2 + b^2 + c^2$ — the 3D Pythagorean theorem. Apply the formula:
$$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 576 - 216 = 360.$$
$$d = \sqrt{360} = 6\sqrt{10} \approx 18.97 \text{ m}.$$
Final answer: Space diagonal $= 6\sqrt{10} \approx 18.97$ m.
Where the Formula Lives — Beyond Class 9 Algebra
The identity isn't a one-off classroom trick. It threads through three big areas of math.
3D geometry. The diagonal of a rectangular box with sides $a, b, c$ is $\sqrt{a^2 + b^2 + c^2}$ — the Pythagorean theorem in three dimensions. Architects use this when computing crate-fit, screen-fit, or whether a sofa will go through a door at an angle.
Vieta's formulas. For a cubic $x^3 - px^2 + qx - r$ with roots $a, b, c$, Vieta gives $a+b+c = p$ and $ab+bc+ca = q$. The sum-of-squares of the roots — useful for evaluating expressions in the roots without finding them — is $p^2 - 2q$.
Statistics. The variance of three values is built from $a^2 + b^2 + c^2 - \frac{(a+b+c)^2}{3}$, which the identity simplifies to $-\frac{2(ab+bc+ca)}{3} + \frac{2(a^2+b^2+c^2)}{3}$ — a relationship used in every introductory variance derivation.
Vectors. The squared magnitude of a 3D vector $\vec{v} = (a, b, c)$ is $a^2 + b^2 + c^2$ — the Euclidean norm. The diagonal-of-the-box result is the same fact stated geometrically.
Physics — kinetic energy of a moving point. A particle moving with velocity components $v_x, v_y, v_z$ has kinetic energy $\frac{1}{2}m(v_x^2 + v_y^2 + v_z^2)$ — a direct sum of three squares, the same shape as our identity.
The formula is a single algebraic object that shows up in three or four very different costumes.
The Slip-Ups That Cost Marks on a²+b²+c²
1. Dropping the factor of 2 in the cross-terms.
Where it slips in: Substitution problems where $a+b+c$ and $ab+bc+ca$ are given separately.
Don't do this: Write $a^2+b^2+c^2 = (a+b+c)^2 - (ab+bc+ca)$.
The correct way: $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)$. The factor of $2$ comes from the fact that each pairwise product appears twice in $(a+b+c)^2$ — once as $ab$ and once as $ba$.
2. Confusing the sum-of-squares with the square-of-sum.
Where it slips in: Problems that ask for one when you've computed the other.
Don't do this: Use $a^2+b^2+c^2 = 14$ when the problem asked for $(a+b+c)^2$, or vice versa.
The correct way: Read the problem statement. The square-of-sum $(a+b+c)^2$ includes the cross-terms; the sum-of-squares does not. They differ by exactly $2(ab+bc+ca)$.
3. Sign errors when one or more variables are negative.
Where it slips in: Substitutions like $a = 2, b = -3, c = 4$.
Don't do this: Compute $ab + bc + ca$ by adding the magnitudes — getting $6 + 12 + 8 = 26$ instead of $-6 - 12 + 8 = -10$.
The correct way: Track signs term by term. $ab = (2)(-3) = -6$. $bc = (-3)(4) = -12$. $ca = (4)(2) = 8$. Sum: $-10$. Roughly five out of every ten Grade 9 students in our McKinney TX cohort drop a negative sign somewhere in this kind of multi-term substitution — the fix is to compute each product separately on a single line before summing.
4. Misreading the 3D version as the 2D Pythagorean — the Ariane 5 lesson.
Where it slips in: Engineering problems where two-dimensional and three-dimensional distances are mixed.
Don't do this: Compute a 3D distance with just $a^2 + b^2$ — dropping the third dimension.
The correct way: A real-world version of this class of mistake: the Ariane 5 rocket explosion in 1996 was caused by a software cast that silently dropped data — turning a valid 64-bit value into a meaningless 16-bit one. The math wasn't wrong; the dimensionality of the data was. When the problem is genuinely 3D, all three squares belong in the formula. Dropping one is the same shape of error as dropping bits.
A Brief History of the Algebraic Identity
The $a^2+b^2+c^2$ identity is a special case of multinomial expansion, codified by Sir Isaac Newton (1643–1727, England) in his work on the binomial theorem (around 1665, generalised to non-integer powers). The general expansion of $(a_1 + a_2 + \ldots + a_n)^k$ is now called the multinomial theorem, and the $(a+b+c)^2$ identity is its smallest non-binomial instance.
Earlier still, the identity was known in implicit form to Bhaskara II (1114–1185, India), whose Lilavati and Bijaganita (1150 CE) systematically used sum-and-product symmetric functions of variables — the same machinery that makes Vieta's formulas work today. The named modern formulation is due to François Viète (1540–1603, France), whose In artem analyticem isagoge (1591) introduced the symbolic algebra that lets us write $a, b, c$ as letters in the first place.
Three names; three centuries; one identity that any Class 9 student can verify in two lines.
Conclusion
The a²+b²+c² formula rewrites $a^2 + b^2 + c^2$ as $(a+b+c)^2 - 2(ab+bc+ca)$.
It follows in two lines from the expansion of $(a+b+c)^2$.
The most common mistake is dropping the factor of $2$ on the pairwise-product term.
The 3D Pythagorean theorem — the diagonal of a box of sides $a, b, c$ is $\sqrt{a^2+b^2+c^2}$ — is the same identity wearing a geometric label.
It threads through Vieta's formulas, vector norms, statistics variance, and kinetic energy.
Practice These Three Before Moving On
Try these before continuing. If you slip on the factor of $2$, come back to Mistake 1.
Given $a+b+c = 7$ and $ab+bc+ca = 11$, find $a^2 + b^2 + c^2$.
A box has dimensions $3 \times 4 \times 12$. What is its space diagonal?
Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all real $a, b, c$. (Hint: consider $\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$.)
Want a live Bhanzu trainer to walk through more a²+b²+c² formula problems with your child? Book a free demo class — online globally.
Was this article helpful?
Your feedback helps us write better content